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LABORATORY  MANUAL  OF 
GENERAL  CHEMISTRY 


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In  this  manual  the  system  of  Qualitative 

Analysis  is  gradually  developed  in. 

its  elementary  form. 

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THE   CALCULATIONS 

OF 

GENERAL  CHEMISTRY 

WITH 

DEFINITIONS,  EXPLANATIONS,  AND 
PROBLEMS 

BY 

WILLIAM  J.  HALE,  PH.D. 

IMRECTOR,  ORGANIC  CHEMICAL  RESEARCH 
DOW  CHEMICAL  COMPANY 

FOURTH    EDITION,  REVISED 

NINTH  THOUSAND 


NEW  YORK 

D.  VAN  NOSTRAND  COMPANY 

8  WARREN  STREET 
1921 


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if**/ 

BIOLOGY 
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BY 

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Stanbope  ipreaa 

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To  THE  MEMORY  OF 

barker 


LATB    PROFESSOR    OF   ORGANIC    CHEMISTRY   IN 

HARVARD   UNIVERSITY 
IN    APPRECIATION    OF    HIS    MOST    INSPIRING   INFLUENCE 

THIS    BOOK 
IS   GRATEFULLY    DEDICATED 


573639 


PREFACE  TO   THE  FIRST  EDITION 


THE  justification  for  a  book  of  this  character  is  to  be 
found  in  the  realization  that  many  of  the  mathematical 
applications  of  our  fundamental  conceptions  in  chemistry, 
even  upon  the  most  elementary  points,  remain  uncompre- 
hended  by  students  several  years  advanced  in  the  study  of 
the  science.  The  great  attention  given  to  the  study  of 
Physical  Chemistry,  and  the  far-reaching  importance 
attached  to  the  interpretation  of  chemical  phenomena  in 
the  light  of  modern  theories,  make  it  absolutely  essential 
that  more  time  be  given  in  the  college  courses  on  General 
Chemistry  to  these  mathematical  demonstrations. 

Unless  a  clear  and  concise  exposition  of  the  methods  of 
calculation  is  presented  in  the  very  first  course  of  college 
work,  the  progress  of  the  student  is  greatly  hindered.  On 
the  other  hand,  too  great  a  mass  of  mathematical  data 
may  impede  rather  than  promote  this  progress.  The  aim, 
therefore,  has  been  to  limit  these  calculations  to  those 
subjects  generally  regarded  as  fundamental,  and  in  which 
the  student  should  receive  the  most  complete  and  thorough 
drill.  Such  may  rightfully  constitute  our  "Arithmetic  of 
Chemistry."  Of  the  many  important  subjects  omitted, 
the  greater  number  possess  a  theoretical  bearing  which 
brings  them  more  properly  into  those  courses  following 
General  and  Analytical  Chemistry  or  into  the  study  of 
Physical  Chemistry  itself. 

The  absence  of  mathematical  training  as  a  basis  for  the 
study  of  chemistry  constitutes  a  widely  prevailing  defect 
in  the  education  of  chemists  at  the  present  day.  The 

vii 


Vlll  PREFACE 

appearance  of  works  on  General  Chemistry  from  the  stand- 
point of  Physical  Chemistry,  with  their  extensive  adoption 
in  our  American  colleges  and  universities,  is  strongly  indic- 
ative of  the  mathematical  trend  in  modern  chemical 
instruction.  Under  this  influence  the  presentation  of  the 
methods  of  chemical  calculations  in  their  simplest  possible 
form  should  facilitate  and  extend  the  use  of  these  modern 
texts  and  at  the  same  time  operate  favorably  in  the  eradi- 
cation of  former  defects.  Primarily,  however,  this  ele- 
mentary presentation  is  intended  to  accompany  the 
laboratory  work  in  General  Chemistry  and  to  help  the 
student  to  a  better  understanding  of  the  many  possible 
conditions  that  may  arise  in  the  course  of  experimental 
work.  The  ideal  plan,  therefore,  should  be  to  associate 
these  problems  and  their  solutions  with  the  actual  labora- 
tory practice.  In  the  study  of  the  last  two,  and  possibly 
three,  chapters  only  a  very  few  correlated  experiments 
may  be  found  to  fit  properly  into  the  laboratory  courses 
on  General  Chemistry;  but  the  important  bearing  of  the 
topics  here  outlined  upon  the  thorough  grounding  of  the 
student's  ideas  of  chemical  reactions  will  make  it  impera- 
tive that  a  number  of  the  more  simple  illustrations  be 
given  by  way  of  lecture  experiments.  Especially  is  this 
true  of  those  reactions  relating  to  combinations  between 
gases  by  volume. 

For  the  greater  part  of  material  presented  no  originality 
is  claimed,  but  gratitude  is  freely  expressed  to  the  authors 
of  the  many  valuable  and  admirable  treatises  on  Stoi- 
chiometry.  In  the  manner  of  presentation  a  somewhat 
different  plan  has  been  followed  from  that  usually  found 
in  books  of  this  nature.  This  consists  in  a  gradual  intro- 
duction of  each  new  condition  properly  falling  under  the 
consideration  of  some  one  subject,  and  the  final  develop- 
ment of  the  subject  in  its  entirety  from  all  the  conditions 


PREFACE  IX 

thus  considered.  By  this  method  it  is  thought  to  obviate 
the  dogmatic  manner  of  presenting  mathematical  facts 
and  to  make  the  student  realize  immediately  the  connec- 
tion between  all  of  our  fundamental  laws  of  energy  and 
matter.  With  this  train  of  thought  connecting  each  and 
every  point  to  that  preceding,  it  is  believed  that  the 
student  will  gain  a  better  and  firmer  hold  upon  the  entire 
subject-matter  rather  than  upon  the  isolated  points. 

This  book  in  itself  comprises  no  more  than  is  presented 
by  the  author  to  students  in  their  first  year  of  chemistry 
at  the  university.  The  problems  at  the  end  of  each  chap- 
ter serve  as  a  guide  and  basis  of  selection  for  examinations 
throughout  the  course.  Great  emphasis,  in  fact,  is  laid 
upon  this  side  of  the  student's  development  as  the  highest 
and  most  beneficial  for  his  future  advancement. 

In  general  the  field  of  use  for  a  book  of  this  type  will  be 
among  colleges  and  universities.  It  is,  however,  sincerely 
hoped  that  at  least  the  first  portion  of  the  book  may  have 
a  helpful  and  profitable  bearing  upon  the  instruction  of 
chemistry  in  the  high  schools. 

For  the  many  valuable  suggestions  and  the  interest  with 
which  criticism  has  been  freely  extended  in  the  revision  of 
the  manuscript,  the  author  expresses  his  sincere  apprecia- 
tion and  lasting  indebtedness  to  Professor  Charles  Loring 
Jackson  of  Harvard  University,  Professor  Alexander  Smith 
of  the  University  of  Chicago,  and  Professor  S.  Lawrence 
Bigelow  of  the  University  of  Michigan. 

WILLIAM  J.   HALE. 
ANN  ARBOR,  May,  1909. 


PEEFACE   TO   THE   SECOND   AND   THIRD 
EDITIONS 


WITH  a  new  and  larger  edition  it  has  seemed  best  to 
make  all  possible  minor  changes  in  both  text  and  problems. 
The  work  of  revision  has  been  greatly  facilitated  through 
the  kind  suggestions  of  those  who  have  found  the  book  help- 
ful, and  of  others  who  have  shown  especial  interest  in 
this  exposition  of  the  subject.  For  all  of  which  grateful 
appreciation  is  heartily  expressed. 

THE  AUTHOR. 

ANN  ARBOR,  January,  1910. 
October,  1911. 


CONTENTS. 


CHAPTER  PAGE 

*J  I.    UNITS  OF  MEASUREMENT 1 

II.   DENSITY  AND  SPECIFIC  GRAVITY  5 

III.   THE  EFFECT  OF  PRESSURE  UPON  GASES 9 

J    IV.   THE  EFFECT  OF  TEMPERATURE  UPON  GASES 15 

*    V.   THE  COMBINED  EFFECT  OF  PRESSURE  AND  TEM- 
PERATURE UPON  GASES.     PARTIAL  PRESSURES  ..  21 
VI.   AVOGADRO'S  HYPOTHESIS  AND  SOME  OF    ITS  AP- 
PLICATIONS   33 

VII.   THE  LAW  OF  DEFINITE  PROPORTIONS 46 

VIII.   THE  DERIVATION  OF  CHEMICAL  FORMULAE 59 

IX.   CALCULATIONS  DEPENDING  UPON  CHEMICAL  EQUA- 
TIONS      82 

X.   NORMAL  SOLUTIONS 106 

XI.   COMBINATIONS  BETWEEN  GASES  BY  VOLUME 122 

XII.   COMPLEX  EQUATIONS 148 

APPENDIX 167 

INDEX 173 


CHEMICAL    CALCULATIONS. 


CHAPTER  I. 
UNITS   OF   MEASUREMENT. 

THE  fundamental  units  employed  in  chemical  calcula- 
tions will  be  defined  at  the  outset  to  insure  their  clear  and 
consistent  use. 

Time.  —  As  the  unit  of  time  we  have  adopted  the 
second,  g^i^  Par^  of  the  mean  solar  day,  or  that  period 
elapsing  between  the  successive  daily  transits  of  the  sun 
across  a  meridian. 

Distance  or  Length.  —  The  meter  as  the  standard  of 
length  is  fixed  as  that  distance  between  two  marks  on  a 
platinum-iridium  bar  preserved  in  Paris  when  this  bar  is 
at  the  temperature  of  0°  C.  It  is  approximately  one 
forty-millionth  of  a  meridian,  or  one  ten-millionth  of  the 
earth's  quadrant  from  pole  to  equator.  Larger  values 
in  multiples  of  the  meter  by  ten,  one  hundred  and  one 
thousand  are  named  by  use  of  the  Greek  prefixes,  —  deca- 
meter, hectometer  and  kilometer  (km.),  —  while  submul- 
tiples  by  the  same  values  receive  names  from  the  Latin 
prefixes,  —  decimeter  (dcm.),  centimeter  (cm.)  and  milli- 
meter (mm.).  The  one  one-hundredth  part  of  the  meter, 
the  centimeter,  is  usually  defined  as  the  unit  of  length. 
As  still  smaller  values  we  employ  the  micron  (/*),  the  one 
one-thousandth  of  a  millimeter,  and  the  millimicron  (///(), 
the  one  one-thousandth  of  a  micron. 

1 


c ,  .CHEMICAL4  CALCULATIONS 


Volume  and  Mass.  —  The  cubic  decimeter  or  liter  is 
the  standard  of  volume.  The  standard  of  mass  is  the 
kilogram,  or  a  mass  of  platinum-indium  in  block  form, 
preserved  in  Paris,  and  originally  intended  to  have  the 
same  mass  as  a  cubic  decimeter  of  water  at  its  greatest 
density,  4°  C.  It  is,  however,  slightly  less  than  a  cubic 
decimeter  of  water  which  at  4°  C.  weighs  1.000013  kilo- 
grams. Both  the  liter  and  kilogram  are  somewhat  large 
for  general  scientific  purposes.  It  is  customary,  therefore, 
to  use  the  one  one-thousandth  part  of  each,  —  the  cubic 
centimeter  (cm.3  or  c.c.)  as  the  unit  of  volume  and  the 
gram  (g.)  as  the  unit  of  mass.  The  mass  of  1  c.c.  of  water 
at  4°  C.  is  considered  as  1  gram.  The  slight  discrepancy 
between  this  value  and  the  true  one  need  be  considered 
only  in  the  most  exact  calculations. 

Through  a  combination  of  the  fundamental  units  just 
mentioned  we  arrive  at  a  standard  system  in  terms  of 
which  so  many  of  our  important  units  of  measurement 
may  be  defined.  This  system  of  units,  known  as  the 
ceritimeter-gram-second  (C.G.S.)  system,  has  met  with 
universal  adoption  throughout  the  scientific  world.  Its 
applications  may  be  noted  in  the  following  paragraph. 

Force  and  Energy.  —  When  a  body  moves  at  a  uniform 
rate  through  a  unit  of  distance, (the  centimeter^  in  a  unit 
of  time,  the  second,  it  is  said  to  have  a  unit  of  velocity 
(abbreviated  cm. /sec.).  When  the  change  in  velocity 
during  one  second  is  one  centimeter  per  second  we  have 
a  unit  of  acceleration.  That  force  which  will  give  to  unit 
mass  unit  acceleration  is  called  the  unit  of  force  or  (j/yne. 
Now  the  same  force  operating  upon  different  bodies  does 
not  produce  in  each  the  same  acceleration,  When,  how- 
ever, the  same  acceleration  is  attained  in  the  several 
cases  the  bodies  must  have  equal  masses.  The  work 
done  by  the  force  of  one  dyne  in  producing  a  displace- 


UNITS  OF  MEASUREMENT  3 

ment  in  its  own  direction  of  one  centimeter  is  called  the 
unit  of  work  or  erg.  When  a  body  gains  or  loses  energy 
the  amount  of  energy  may  be  measured  in  units  of  work, 
and  when  the  work  done  upon  a  body  imparts  to  it  a 
certain  velocity,  the  body,  by  virtue  of  this,  is  said  to 
possess  kinetic  energy.  The  amount  of  energy  so  pos- 

MV2 
sessed  may  be  expressed  by  the  formula,  E  =  — - — ,  in 

which  E  represents  the  energy,  M  the  mass  and  V  the 
velocity.  Mass,  then,  is  in  itself  but  a  "  measure  of  the 
kinetic  energy  which  a  body  possesses  when  it  has  a 
definite  velocity."  —  Ostwald. 


Standards  of  Temperature  and  Pressure. 

The  accuracy  with  which  measurements  are  conducted 
requires  certain  definite  and  constant  conditions  of  tem- 
perature and  pressure,  the  effect  of  changes  in  which  will 
be  studied  in  a  later  chapter.  The  standards  adopted  with 
these  factors  will  receive  only  brief  mention  at  this 
point. 

Temperature.  —  The  freezing-point  of  pure  water  is 
taken  as  the  normal  temperature  and  made  0°  upon  the 
Celsius  or  centigrade  thermometer.  The  boiling-point  of 
pure  water  at  760  mm.  pressure  is  registered  at  100°  on 
this  scale,  and  the  intervening  range  of  temperature 
between  the  freezing-  and  boiling-points  of  water  is  di- 
vided into  one  hundred  equal  parts  or  degrees  centigrade. 
As  pressure  exerts  a  considerable  influence  upon  the 
boiling-points  of  liquids,  all  temperatures  are  referred  to 
the  normal  condition  of  pressure. 

Pressure.  —  The  standard  condition  of  pressure,  the 
normal  pressure,  is  taken  as  that  pressure  which  the 
atmosphere  exerts  at  sea  level  in  a  latitude  of  45°.  This 


4  CHEMICAL   CALCULATIONS 

pressure  is  sufficient  to  sustain  a  column  of  mercury,  at 
0°  C.,  76  cm.  (760  mm.)  in  height.  In  a  column  of  this 
height  and  1  sq.  cm.  cross  section  there  are  exactly  76  c.c. 
of  mercury,  and  since,  volume  for  volume,  mercury  is 
13.596  times  as  heavy  as  water  and  1  c.c.  of  it  therefore 
weighs  this  number  of  grams,  we  have  at  once  the  value 
76  X  13.596  =  1033.2  grams  as  the  weight  of  the  atmos- 
phere per  square  centimeter.  By  reference  to  a  barometer 
the  actual  weight  of  the  atmosphere  in  millimeters  of 
mercury  is  observed  under  the  various  conditions.  These 
barometric  readings  are  usually  made  at  other  tempera- 
tures than  0°  C. ;  the  correct  readings,  therefore,  for  milli- 
meters of  mercury  at  0°  C.  must  be  calculated  by  use  of  a 
proper  table  showing  the  expansion  of  mercury  with  tem- 
perature (Appendix  I).  Gases  measured  in  vessels  in- 
verted over  a  liquid  will  of  course  have  the  same  pressure 
as  the  atmosphere  (recorded  by  the  barometer)  when  the 
levels  of  the  liquid  in  the  vessel  and  outside  of  it  are  equal. 
The  temperature  of  the  gases  and  liquid  should  of  course 
be  alike. 


CHAPTER  II. 
DENSITY   AND   SPECIFIC   GRAVITY. 

THE  term  density  is  used  to  designate  the  mass  in  unit 
of  volume.  As  this  is  a  constant  for  any  substance  under 
given  conditions  of  temperature  and  pressure,  we  may  call 
it  the  absolute  density.  It  is  expressed  simply  as  D  =  M/  V, 
where  D,  M  and  V  represent  respectively  density,  mass 
and  volume.  When  the  absolute  density  of  one  substance 
is  referred  to  that  of  another,  i.e.,  a  relation  between  these 
two  densities  determined  upon  the  basis  of  one  of  them  as 
a  standard,  we  have  a  value  known  as  the  specific  or  rela- 
tive density,  or  what  has  been  more  commonly  called  the 
specific  gravity.  The  relative  density  of  a  substance  is  not 
affected  by  the  force  of  gravity,  since  any  change  in  this 
force  will  operate  equally  upon  the  absolute  densities  of 
all  substances  as  well  as  upon  the  standard.  The  relative 
density  or  specific  gravity  may  be  defined  simply  as  the 
ratio  between  the  weight  of  any  substance  and  that  of  an 
equal  volume  of  the  standard  substance  when  this  latter 
is  considered  as  unity.  As  a  standard  for  the  specific 
gravity  of  liquids  and  solids,  the  most  natural  choice  is 
distilled  water,  which  is  considered  at  the  point  of  its 
greatest  density,  namely  4°  C.  Determinations  made  at 
other  than  this  temperature  (e.g.  15°)  are  usually  referred 
to  the  density  of  water  at  4°  C.,  and  expressed  thus: 
sp.  gr.  at  lo°/4°. 

In  the  case  of  gases,  the  values  obtained  for  the  absolute 
densities,  the  number  of  units  of  mass  per  unit  of  volume, 
are  exceedingly  small.  By  reason  of  this  it  is  more  cus- 

5 


6  CHEMICAL   CALCULATIONS 

ternary  to  multiply  these  values  by  1000  or,  what  is  the 
same  thing,  to  determine  the  number  of  units  of  mass  in 
1  liter  instead  of  1  c.c.  Though  the  densities  of  gases 
referred  to  some  standard  such  as  air  are  known  as  relative 
densities,  the  term  vapor  density  meets  with  constant  use 
when  the  relative  density  is  determined  for  a  liquid  or 
solid  in  the  state  of  vapor.  Since  the  composition  of  the 
atmosphere  is  somewhat  variable,  the  relative  densities 
upon  this  standard  cannot  possess  much  accuracy.  The 
employment  of  a  pure  gaseous  substance,  such  as  hydrogen, 
gives  much  better  results.  As  the  lightest  gas,  all  other 
densities  referred  to  it  assume  values  greater  than  unity. 
The  absolute  density  of  hydrogen,  the  weight  in  grams  of 
1  liter  at  0°  and  760  mm.,  is  0.08987.  The  weight  of  an 
equal  volume,  1  liter,  of  oxygen  under  the  same  conditions 
is  1.429  grams;  hence  with  reference  to  hydrogen  as  unity, 
oxygen  will  be  found  to  have  a  relative  density  of  15.90, 
i.e.,  1.429  is  15.90  times  0.08987,  or,  by  simple  proportion: 

0.08987:  1.429  =  1:  x,  or  x  =  15.90. 

The  Oxygen  Standard.  —  The  intimate  relation  (to  be 
noted  later)  between  the  molecular  weight  of  a  substance 
and  the  volume  which  this  weight  occupies  in  the  state  of 
vapor  has  led  us  to  refer  all  densities  of  gaseous  substances 
to  the  density  of  that  substance  which  is  to  serve  as  the 
basis  for  molecular  weights.  Formerly  hydrogen,  as  the 
lightest  substance,  served  this  purpose,  and  consequently 
the  close  relationship  between  densities  and  molecular 
weights  was  apparent. 

In  recent  times  oxygen,  with  the  value  32,  has  been 
adopted  as  the  basis  of  molecular  weights  by  reason  of 
the  great  importance  of  this  element  in  its  numerous 
combinations  with  other  elements  and  for  reasons  that 
be  made  clear  after  further  considerations.  The 


DENSITY   AND   SPECIFIC   GRAVITY  7 

absolute  density  of  oxygen,  therefore,  is  now  taken  as  the 
standard  of  gas  densities  and  made  equal  to  unity.  Upon 
this  basis  the  close  relationship  between  the  densities  of 
gases  and  their  corresponding  molecular  weights  will  be 
as  well  attained  as  originally,  when  hydrogen  was  the 
standard.  The  relative  densities  of  a  number  of  sub- 
stances, however,  will  fall  below  unity.  Thus  the  relative 
density  of  hydrogen  becomes  on  this  scale  0.0629,  a  value 
easily  derived  from  the  simple  proportion: 

1.429  :  0.08987  =  I:  x. 

In  order  to  avoid  these  small  values  some  chemists  prefer 
to  assign  to  oxygen  the  absolute  density  of  16  instead  of 
unity.  From  the  relations  already  noted  between  oxygen 
on  the  one  hand  and  hydrogen,  as  the  lightest  substance, 
on  the  other,  we  readily  see  that  the  relative  densities  of 
•all  other  substances  are  thus  brought  to  values  greater 
than  unity.  By  the  adoption  of  32  instead  of  16,  as  this 
basis  of  densities,  the  relative  densities  may  be  made  to 
coincide  at  once  with  the  molecular  weights.* 

Under  all  conditions  the  measure  of  the  volumes  of 
gases  is  conducted  at  definite  temperatures  and  pressures. 
The  influence  of  change  in  these  factors  will  be  discussed 
in  the  succeeding  chapters.  Whatever  the  volume  of  a 
gas  may  be,  its  relative  density  will  be  determined  from 
the  ratio  of  the  observed  weight  per  volume  to  the  weight 

*  Strictly  speaking,  an  exact  agreement  on  this  basis  between 
relative  densities  and  molecular  weights  is  found  true  only  in  the 
case  of  perfect  gases  (c/.  footnote,  p.  34).  In  the  study  of  actual 
weights  of  equal  volumes  of  gases  it  seems  preferable,  therefore, 
to  base  the  comparison  upon  the  weight  of  the  standard  as  unity. 
The  significance  of  the  theoretical  or  calculated  values,  upon  the 
basis  of  32  as  the  chemical  unit,  will  be  noted  in  succeeding 
chapters. 


8  CHEMICAL  CALCULATIONS 

of  an  equal  volume  of  oxygen  under  like  conditions,  when 
this  ratio  is  brought  over  to  the  basis  of  oxygen  as  unity. 
Example  1. —  The  actual  weight  of  1  liter  of  carbon  dioxide,  at 
standard  conditions,  is  1.977  grams.  One  liter  of  oxygen  weighs 
1.429  grams.  What  is  the  relative  density  of  carbon  dioxide? 

The  ratio  between  these  two  densities  is  now  to  be 
referred  to  the  density  of  oxygen  as  unity.     That  is,  the 

1  429  1 

ratio    '        is  to  be  made  equal  to  the  ratio  - ,  where  x  is 
Ji.y  ii  x 

the  relative  density  of  carbon  dioxide.     We  thus  have 

1.429      1  . 

Q7,7  =  - ,  or,  as  more  commonly  expressed, 
i.  y  / 1      x 

1.429  :  1.977  =  I:  x. 
From  which  x  is  found  to  be  1.384. 

PROBLEMS. 

1.  A  piece  of  metal  weighing  30  grams  displaced  20  c.c.  of 
water  (i.e.,  its  own  volume).     What  is  the  relative  density  of 
this  metal  referred  to  water?  Ans.    1.5. 

2.  A  vessel  weighing  6.448  grams  weighed  7.963  grams  when 
filled  with  water  and  8.266  grams  when  filled  with  a  salt  solu- 
tion.   What  is  the  relative  density  of  this  solution  referred  to 
water?     (The  temperature  constant  throughout.) 

Ans.    1.2. 

3.  The  weight  of  1  liter  of  aqueous  vapor  at  0°  and  760  mm. 
is  0.8045  gram.    What  is  its  relative  density?      .4ns.  0.5630. 

4.  Calculate  the  relative  density  of  mercury  vapor,  1  liter  of 
which  at  standard  conditions  (0°  and  760  mm.)  weighs  8.87 
grams.  Ans.    6.207. 

5.  Calculate  the  relative  density  of  hydrogen  chloride,  5  liters 
of  which  under  standard  conditions  weigh  8.205  grams. 

Ans.    1.148. 

6.  Calculate  the  relative  density  of  chlorine,  100  c.c.  of  which 
at  standard  conditions  weigh  0.322  gram.  Ans.    2.253. 


CHAPTER  III. 

THE    EFFECT    OF    PRESSURE    UP0N    GASES. 
THE    LAW   OF   BOYLE. 

The  Relation  of  Volume  to  Pressure.  —  When  a-  gas  is 
subjected  to  increase  of  pressure  the  volume  decreases. 
On  release  of  the  pressure  the  volume  increases,  regain- 
ing its  former  volume  only  at  the  original  pressure.  The 
Law  of  Boyle  states  this  as  follows:  At  constant  tem- 
perature the  product  of  the  volume  of  a  gas  (V)  by  its 
pressure  (P)  is  a  constant,  or  P  X  V  =  K.  Thus  a 
volume  of  gas  (1)  at  a  pressure  of  one  atmosphere  (1) 
will  be  reduced  when  under  a  pressure  of  two  atmos- 
pheres to  one-half  of  its  original  volume,  or,  in  the  equa- 
tion P  X  V  =  K,  the  values  1X1  =  1  will  become 
2  X  -J-  =  1.  This  constancy,  however,  applies  to  all 
gases,  and  hence  the  product  PV  in  one  case  will  be  equal 
to  the  product  P'V  for  any  other  case  when  either  P  or  V 
or  both  change,  the  temperature  considered  constant 
throughout;  hence  the  equation  PV  =  P'V. 

In  dividing  this  equation,  PV  =  P'V7,  through  by  the 
quantity  PV7  and  canceling  like  terms  in  numerator  and 

PV        P'V'     V       P' 
denominator,  we  obtain  p^-  =  p^-or^  =  —  .  The  two 

fractional  quantities  may  be  expressed  also  by  the  propor- 
tion V  :  V  =  P'  :  P.  This  is  exactly  what  is  indicated 
above,  where  a  volume  of  gas  under  a  definite  pressure 
will  have  its  volume  halved  under  double  this  pressure, 
etc.,  or  at  constant  temperature  the  volume  of  a  gas  is 
inversely  proportional  to  its  pressure  (a  general  form  for 

9 


10  CHEMICAL   CALCULATIONS 

the  statement  of  Boyle's  Law).  Strictly  speaking,  the 
Law  of  Boyle  holds  true  only  for  a  perfect  gas.  For  prac- 
tical purposes  the  law  gives  a  sufficiently  accurate  means 
of  studying  volume  and  pressure  relations. 

In  order  to  ascertain  what  a  given  volume  of  gas  may 
become  at  some  new  pressure,  the  temperature  constant 
throughout,  we  have  only  to  apply  the  formula  PV  =  P'V, 
in  which  we  may  designate  the  new  values  by  the  prime 
marks. 

Example  2.  —  200  c.c.  of  hydrogen  measured  at  750  mm. 
pressure  will  occupy  what  volume  at  the  standard  pressure 
(760  mm.),  temperature  a  constant? 

PV  =  P;Y(  or  V  =  V  (P/P')-  Substituting  here  the 
respective  values  P'  =  760,  P  =  750  and  V  =  200,  we 
obtain  the  expression  V'  =  200  X  750/760,  which  is  easily 
solved  to  a  value  of  197.3. 

In  general  terms,  we  shall  say  that  a  change  in  the  volume 
of  any  gas  by  change  in  pressure  is  readily  calculated  from 
the  values  given,  by  multiplying  the  volume  by  the  fraction 
formed  from  the  original  pressure  as  numerator  and  new 
pressure  as  denominator.  If  this  new  pressure  is  greater 
than  the  original,  then  the  fraction  will  have  a  value  less 
than  unity  and  the  new  volume  (the  product  of  fraction 
by  original  volume)  will  be  proportionately  less;  con- 
versely, if  the  new  pressure  is  smaller  than  the  original 
pressure,  the  fraction  (formed  from  the  pressures)  be- 
comes larger  than  unity  in  value,  and  the  new  volume 
must  increase  over  that  of  the  original.  It  comes,  there- 
fore, to  the  same  end  if  one  ask  himself  the  question: 
"  Is  the  new  pressure  greater  or  less  than  the  original?  " 
If  greater,  then  the  fraction  formed  from  the  pressures 
must  be  so  arranged  that  the  new  volume  of  the  gas  will 
be  smaller,  i.e.,  the  fraction  must  have  as  its  numerator 


THE  EFFECT  OF  PRESSURE  UPON  GASES     11 

the  smaller  of  the  two  pressure  values,  and  as  its  denomi- 
nator the  larger.  This  fraction  less  than  unity  in  value 
will,  as  was  just  shown,  reduce  the  volume  which  it  mul- 
tiplies in  the  correct  proportion.  If  the  new  pressure  is 
smaller  than  the  original,  the  diminution  in  pressure  must 
be  accompanied  by  an  expansion  in  volume,  and,  accord- 
ingly, the  fraction  formed  from  the  pressures,  which  we 
may  conveniently  call  the  pressure-fraction,  takes  the 
larger  value  in  the  numerator  and  smaller  value  in  the 
denominator.  From  the  product  of  this  fraction,  greater 
than  unity,  by  the  original  volume  we  obtain  the  new  and 
correspondingly  increased  volume. 

The  Relation  of  Pressure  to  Volume.  —  Just  as  change 
in  pressure  produces  a  change  in  volume,  so  also  a  change 
in  volume  will  produce  a  change  in  the  pressure  of  a  gas. 
The  effect  of  this  change  in  volume  upon  the  pressure  is 
readily  determined  from  the  equation  PV  =  P'V,  and 
takes  the  form  of  expression: 

P'  =  p  (V/V). 

As  the  new  volume  V  increases  or  decreases  with  reference 
to  the  original,  the  volume-fraction  becomes  respectively 
less  or  greater  than  unity  in  value. 

Examples. — A  volume  of  gas  measuring  100  c.c.  at  750 
mm.  was  expanded  to  500  c.c.  at  constant  temperature.  What 
was  the  pressure  of  the  gas  at  this  final  volume? 

In  order  to  arrange  the  fraction  (formed  from  the 
volumes)  in  such  a  way  as  to  give  a  value  less  than  unity 
and  one  that  will  reduce  the  given  pressure  in  the  ratio 
indicated  by  Boyle's  Law,  it  is  necessary  here  to  place  the 
smaller  number  in  the  numerator.  The  product  of  this 
volume-fraction  by  the  original  pressure  P,  or  750  mm., 
gives  the  new  pressure  sought:  750  X  100/500  =  150. 


l\ 


12  CHEMICAL   CALCULATIONS 

The  Relation  of  Density  to  Pressure.  —  The  absolute 
density  of  a  gas,  the  mass  per  liter,  may  be  seen  now  to 
vary  with  the  pressure  to  which  the  gas  is  subjected. 
With  reference  to  the  formula  D  =  M/V,  we  are  aware 
that  the  weight  or  mass  is  unaffected  by  any  changes 
in  temperature  or  pressure.  The  volume  (V),  however,  is 
altered  by  either  or  both  of  these  influences.  Thus  with 
an  increase  of  pressure  it  decreases  and  hence  the  absolute 
density  must  increase.  In  other  words,  both  the  absolute 
density  and  the  pressure  of  a  definite  quantity  of  gas  vary 
inversely  with  the  volume  when  the  temperature  is  con- 
stant, and  hence  are  directly  proportional  to  each  other. 
Therefore,  in  the  equation  PV  =  P'V  we  may  substitute 
D  for  P  throughout  and  obtain 

DV  =  D'V,  or  D'  =  D  (V/V). 

Example  4.  — 100  c.c.  of  hydrogen,  the  absolute  density  of 
which  (the  weight  of  1  liter)  is  0.08987,  will  have  what  density 
when  expanded  to  200  c.c.  in  volume  at  a  constant  tem- 
perature? 

Here  the  increase  in  volume  means  a  decrease  in  the 
pressure  and  hence  in  the  density.  The  fraction  formed 
from  the  volumes,  the  volume-fraction,  must  be  arranged 
so  as  to  produce  this  decrease  corresponding  to  Boyle's 
Law  and,  therefore,  the  numerator  must  contain  the 
smaller  of  the  two  values.  When  this  quantity  100/200 
is  multiplied  by  the  density  0.08987,  we  obtain  the  new 
density  0.0449. 

This  is  identical  with  saying  that  at  this  new  and  larger 
volume  the  actual  weight  of  the  gas  has  not  changed,  or 
the  original  weight  of  the  100  c.c.,  0.008987  gram,  has 
remained  a  constant.  The  mass  per  volume,  however, 
has  been  changed,  in  that  this  weight  of  gas  is  now  spread 


THE  EFFECT  OF  PRESSURE  UPON  GASES     18 

out  over  twice  the  original  volume.  The  actual  weight 
of  100  c.c.  of  the  rarefied  gas  is  just  one-half  of  the  weight 
of  the  total  volume,  200  c.c.,  or  0.^0449  gram.  As  a 
basis  for  calculations  of  this  nature,  tliat  general  property 
of  homogeneous  substances  is  of  course  understood:  The 
weight  of  any  fractional  part  of  a  volume  bears  the 
same  ratio  to  the  total  weight  as  this  fractional  volume 
itself  does  to  the  total  volume. 

It  should  be  remembered  in  this  connection  that  the 
relative  density  of  a  solid  or  liquid  may  vary  with  the 
temperature  or  pressure  (p.  5),  since  the  effect  of  change 
in  these  factors  upon  the  volume  of  a  substance  and  upon 
that  of  the  standard  chosen  may  not  be  the  same.  The 
relative  density  of  a  gas,  however,  is  constant,  since 
changes  in  these  factors  affect  all  gases  alike. 

PROBLEMS. 

7.  512  c.c.  of  hydrogen  at  a  pressure  of  744  mm.  will  occupy 
what  volume  at  a  pressure  of   790.5  mm.,  the   temperature 
constant?  Ans.  481  c.c. 

8.  240  c.c.  of  gas  a,t  a  pressure  of  740  mm.  were  admitted  into 
an  empty  vessel  of  800  c.c.  capacity.    What  was  the  pressure  of 
the  gas  at  this  new  volume,  temperature  constant? 

Ans.  222mm. 

9.  500  c.c.  of  oxygen,  absolute   density   1.429,  were  com- 
pressed to  a  volume  of  125  c.c.  at  constant  temperature.     What 
was  the  density  of  the  gas  at  this  final  volume?    What  would 
be  the  weight  of  50  c.c.  of  the  compressed  gas? 

Ans.  5.716,  0.2858  gram. 

10.  A  volume  of  gas  measuring  600  c.c.  at  760  mm.  pres- 
sure was  expanded  to  a  volume  of  1000  c.c.  at  constant  tem- 
perature.    What  was  the  final  pressure  of  the  gas,  and  what 
fractional  change  in  its  absolute  density  must  have  followed 
this  expansion?  Ans.   456  mm. 

Bensity  I  of  former  value. 


14  CHEMICAL   CALCULATIONS 

11.  500  c.c.  of  a  gas,  the  absolute  density  of  which  is  6, 
must  be  reduced  to  a  density  of  0.75  at  constant  temperature. 
What  will  be  the  volume  of  this  rarefied  gas?    Calculate  also 
the  weight  of  400  c.c.  of  the  rarefied  gas. 

Ans.  4000  c.c. ;  0.3  gram. 

12.  A  volume  of  gas  weighing  5  grams  was  expanded,  at  a 
constant  temperature,  till  the  pressure  was  reduced  to  one-half 
of  its  former  value.     500  c.c.  of  the  rarefied  gas  weighed  1.25 
grams.     What  was  the  original  volume  of  the  gas?    Calculate 
also  the  original  density  assuming  the   original   observations 
made  at  standard  conditions.  Ans.   1000  c.c. 

Density  5. 


CHAPTER  IV. 

THE  EFFECT  OF  TEMPERATURE  UPON  GASES. 
THE  LAW  OF  CHARLES. 

WITH  pressure  constant  the  volume  of  a  gas  is  found  to 
vary  with  the  temperature.  The  Law  of  Charles  (known 
also  as  Gay-Lussac's  Law),  states  this  as  follows:  Under 
constant  pressure  the  volume  of  a  gas  varies  directly  with 
the  temperature  upon  the  absolute  scale. 

The  Determination  of  the  Coefficient  of  Expansion.  — 
The  standard  condition  of  temperature,  0°  C.,  becomes  on 
the  absolute  scale  273°.  This  is  calculated  from  the  fact 
that,  given  any  volume  of  gas  at  0°  C.,  the  effect  of  cooling 
upon  this  gas  will  bring  about  a  diminution  in  its  volume 
by  1/273  for  each  degree  centigrade  below  the  centigrade 
zero,  consequently  at  —  273°  C.  the  volume  will  have  been 
diminished  by  273/273  of  itself,  or  by  its  entirety.  This 
point,  therefore,  is  regarded  as  the  zero  point  or  lowest 
point  at  which  a  gas  may  possibly  come  under  considera- 
tion. At  or  below  this  point  no  gas  could  exist  as  such. 
Though  this  temperature  has  not  as  yet  been  reached, 
it  follows  that  all  gases  will  be  liquefied  above  it.  A 
volume  of  gas  measuring  273  c.c.  at  0°  C.  would  lose  per 
degree  of  cooling  1/273  of  itself,  or  a  single  unit  of  volume. 
At  one  degree  of  temperature  above  the  absolute  zero 
(designated  as  1°  A.)  the  gas,  having  cooled  through  272° 
and  lost  272/273  of  its  volume  at  0°  C.,  would  consequently 
occupy  but  1  c.c.  At  2°  A.  its  volume  would  be  2  c.c., 
having  lost  here  271/273  of  its  volume  at  0°  C.  At 
273°  A.  its  volume  would  be  again  273  c.c.;  in  fact  through- 

15 


16  CHEMICAL   CALCULATIONS 

out  the  entire  range,  0°  A.  to  273°  A.,  a  unit  of  volume 
corresponds  to  a  unit  of  temperature,  and  the  one  is  there- 
fore directly  proportional  to  the  other,  i.e.,  the  volume 
varies  directly  with  the  temperature.  Upon  passing  the 
point  273°  A.  the  same  principle  holds.  For  each  degree 
of  temperature  there  is  an  expansion  in  volume  corre- 
sponding to  1/273  of  the  volume  at  273°  A.  or  0°  C.  At 
283°  A.  the  volume  of  a  gas  measuring  273  c.c.  at  273°  A. 
will  have  increased  by  10/273  of  this  volume  and  measure 
283  c.c.  The  rate  of  expansion  above  0°  C.  is  identical 
with  that  of  cooling  below  this  temperature.  This  ratio 
between  the  increase  in  volume  per  degree  and  the  total 
volume  of  a  gas  at  0°  C.  — called  the  coefficient  of  expan- 
sion and  often  expressed  decimally  as  0.00367  in  place  of 
the  fraction  1/273  —  was  determined  by  measuring  the 
expansion  of  gases  from  0°  C.  upward  and  this  upon 
the  centigrade  scale;  consequently  the  determination  of 
the  zero  point  on  the  absolute  scale  is  made  in  centigrade 
units,  and  each  degree  of  temperature  centigrade  must 
be  equivalent  to  one  degree  on  the  absolute  scale.  If 
273°  A.  is  the  zero  of  centigrade,  then  283°  A.  is  equal  to 
10°  C.  and  so  on.  In  order  to  preserve  this  direct  pro- 
portionality between  volume  and  temperature  we  have 
only  to  convert  the  centigrade  temperatures  into  their 
corresponding  values  upon  the  absolute  scale.  By  reason 
of  the  equivalence  in  the  units  we  add  273°  to  the  read- 
ings above  0°  C.  and  subtract  readings  below  0°  C.  from 
273°;  the  results  represent  absolute  temperatures  (desig- 
nated by  T  in  contradistinction  to  t  for  the  centigrade 
temperatures). 

With  this  one  volume,  273  c.c.  as  above  chosen,  the 
absolute  temperature  coincides  numerically  throughout 
with  the  volume  in  cubic  centimeters,  and  simplifies  the 
calculations  to  adding  or  subtracting  single  units.  Had 


THE   EFFECT   OF   TEMPERATURE  UPON    GASES      17 

any  other  volume  of  gas  at  0°  C.  been  considered,  the 
same  ratio  between  volume  and  temperature  would  be 
found — a  direct  proportionality — but  the  variations  would 
not  be  in  single  units.  The  equation  which  represents 
this  change  in  volume  V  under  a  definite  change  in  tem- 
perature t  (centigrade)  may  be  simply  expressed  as 
V  =  V  (1  +  at),  where  a  is  the  coefficient  of  expansion 
per  degree  (1/273)  and  t  the  number  of  degrees  centigrade 
through  which  it  operates.  Thus  for  a  definite  rise  in 
temperature,  t,  the  total  expansion  will  be  equal  to  that 
fractional  part  of  the  original  volume  as  is  indicated  by 
the  quantity  at,  i.e.,  (1/273  X  t)  V;  and  any  new  volume 
V  must  be  equal  to  the  original  volume  V  plus  the  total 
expansion : 

V'  =  V  +  (1/273  X  0  Vor 

V  =  V  (1  +  */273)  or 

V  =  V(l  +  at). 

It  is  regarded  as  more  helpful  in  this  work  to  arrange  the 
original  and  new  values  as  members  of  a  simple  propor- 
tion, in  which,  of  course,  only  the  absolute  temperatures 
are  considered:  V7  :  V  =  T'  :  T.  The  application  of  the 
Law  of  Charles  to  various  cases  may  be  illustrated  now 
by  examples. 

The  Relation  of  Volume  to  Temperature. 

Example  5,  —  A  volume  of  gas  measuring  200  c.c.  at  20°  C. 
will  have  what  volume  at  the  standard  condition,  0°  C. 

From  the  direct  proportionality  between  volume  and 
absolute  temperature  we  have 

V  :  V7  =  T  :  T',  or  V  =  V  (T'/T), 

where  T  and  T'  represent,  respectively,  the  original  and 
new  temperatures  expressed  in  absolute  units.  In  the 


18  CHEMICAL   CALCULATIONS 

example  above,  T  =  20°  +  273°  =  293°;  the  new  tem- 
perature is  273°.  The  change  in  volume  of  this  gas  must 
follow  according  to  the  proportion: 

200  :  x  =  293  :  273,  or  x  =  200  (273/293),  or  x  =  186.35 

Here  we  see  the  original  volume  is  to  be  multiplied  by  a 
fraction  formed  from  the  two  temperatures,  the  tempera- 
ture-fraction, and  again  we  may  apply  the  question  as  in 
the  consideration  of  Boyle's  Law:  "Is  the  new  volume  to 
be  smaller  or  greater  than  the  original  volume?"  If 
greater,  then  the  larger  number  (the  degrees  in  absolute 
units)  must  be  set  over  the  smaller  in  order  that  the  frac- 
tional quantity,  when  multiplied  by  the  original  volume, 
may  increase  that  volume  in  the  proportion  indicated  by 
these  temperature  values.  If  the  new  volume  is  to  be 
smaller,  then  the  numerator  must  contain  the  smaller  of 
the  two  numbers  and  the  product  accordingly  will  be 
proportionately  less. 

The  Relation  of  Temperature  to  Volume. 

Example  6.  —  A  volume  of  gas  measuring  100  c.c.  at  17°  C. 
was  expanded  by  warming  to  a  volume  of  250  c.c.  at  constant 
pressure.  To  what  temperature  must  the  gas  have  been  heated 
to  bring  about  this  expansion? 

From  the  Law  of  Charles  we  draw  up  the  equation 
T'  =  T  (V'/V).  In  conformity  with  the  direct  propor- 
tionality between  volume  and  absolute  temperature,  the 
new  temperature  (T')  must  be  greater  than  the  original 
temperature.  The  volume-fraction  thus  arranged  to 
bring  about  this  proportional  increase  when  multiplied 
by  the  original  temperature  (T  =  273°  +  17°  =  290°) 
becomes  250/100,  and  the  final  temperature  accordingly 
is  T'  =  290  (250/100)  or  725°.  725°  A.  is  equal  to  452°  C. 


THE  EFFECT  OF  TEMPERATURE  UPON  GASES   19 

The  Relation  of  Density  to  Temperature.  —  In  con- 
sideration of  the  change  in  absolute  density  under 
change  of  temperature,  with  pressure  a  constant,  we  see 
from  the  formula  D  =  M/V  that,  as  the  volume  increases, 
the  fraction  M/V  (the  expression  for  the  density)  must 
become  less  and  less.  With  the  volume  of  a  gas  —  the 
denominator  of  this  fractional  quantity  —  becoming  less, 
we  obtain  a  larger  quotient  or  a  larger  value  for  the  den- 
sity. Now  as  this  change  in  volume  is  directly  propor- 
tional to  a  change  in  absolute  temperature,  the  density  of 
a  gas  will  decrease  with  an  increase  in  its  absolute  tem- 
perature and  increase  with  a  decrease  in  this  factor,  — 
that  is,  under  constant  pressure  the  absolute  density  of  a 
definite  quantity  of  gas  is  inversely  proportional  to  its 
absolute  temperature  or  D  :  D'  =  T'  :  T. 

Example  7.  —  A  volume  of  hydrogen  at  0°  C.,  with  the  abso- 
lute density  0.08987,  will  have  what  density  when  expanded  by 
heating  to  10°  C.  at  constant  pressure? 

From  the  equation  above  we  have  D'  =  D  (T/T'). 
This  signifies  that  the  density  given  must  be  multiplied 
by  a  fraction  formed  from  the  two  temperatures  (the 
temperature-fraction).  We  should  now  ask  the  question 
whether,  from  the  inverse  proportionality  just  mentioned, 
the  new  density  (D')  will  be  greater  or  less  than  the 
original  density  (D).  Here  the  new  temperature  is 
higher,  consequently  the  new  density  will  be  lower,  there- 
fore the  smaller  number  must  be  placed  in  the  numerator 
and  the  value  for  D'  can  be  readily  calculated: 

D7  -  D  (273/283),  or  0.08987  (273/283),  or  0.0867. 


20  CHEMICAL   CALCULATIONS 

PROBLEMS. 

13.  What  will  be  the  volume  of  250  c.c.  of  hydrogen  meas- 
ured at  30°  when  cooled  to  —  10°  at  constant  pressure? 

Ans.  217  c.c. 

14.  A  rubber  balloon  containing  400  c.c.  of  oxygen  measured 
at  -  20°  is  subjected  to  a  temperature  of  120°.     What  is  the 
increase  in  volume  of  the  balloon,  the  atmospheric  pressure 
constant  throughout?  Ans.  221.3  c.c. 

15.  A  volume  of  gas  measuring  500  c.c.  at  0°  was  expanded 
by  heating  to  600  c.c.,  at  constant  atmospheric  pressure.     What 
was  the  temperature  which  the  gas  attained?         Ans.  54.6°. 

16.  The  absolute  density  of  oxygen  is  1.429.     What  would 
be  the  weight  of  one  liter  of  oxygen  collected  at  40°  and  760  mm. 
pressure?  Ans.  1.246. 

17.  A  volume  of  gas  with  the  absolute  density  4  and  measur- 
ing 250  c.c.  at  0°  was  expanded  by  warming,  under  constant 
pressure,  to  a  volume  of  600  c.c.    What  increase  in  temperature 
was  required,  and  what  would  be  the  weight  of  300  c.c.  of  the 
rarefied  gas?  Ans.  382.2°;  0.5  gram. 


CHAPTER  V. 

THE    COMBINED   EFFECT   OF   TEMPERATURE   AND 
PRESSURE    UPON    GASES.     PARTIAL   PRESSURES. 

WHEN  a  volume  of  gas  is  subjected  to  change  in  both 
temperature  and  pressure  the  relation  of  the  original 
volume  to  the  new  volume  must  be  in  accordance  with 
the  two  principles  already  enunciated: 

V  :  V  =  P  :  P'  and    V'  :  V  =  T'  :  T 

or        V'/V  =  P/P'  and      V'/V  =  T'/T 

or  V  =  V  (P/P')        and          V  =  V  (T'/T). 

Considered  separately,  we  know  that  the  original  volume 
under  change  in  pressure  must  adjust  itself  in  accordance 
with  the  ratio  denoted  by  Boyle's  Law  and  attain  that 
value  which  is  expressed  by  the  product  of  the  pressure- 
fraction  by  the  original  volume,  —  V  (P/P').  The 
operation  of  a  change  in  temperature  will  modify  the 
original  volume  in  accordance  with  the  Law  of  Charles, 
and  the  resulting  value  will  be  expressed,  as  the  formula 
above  indicates,  by  the  product  of  the  original  volume 
by  the  temperature-fraction,  —  V  (T'/T). 

If  the  original  volume,  already  modified  to  accord  with 
a  change  in  pressure,  is  to  come  in  turn  under  the  influence 
of  a  change  in  temperature,  which  is  separate  and  inde- 
pendent of  the  pressure,  we  shall  be  able  to  express  the 
result  by  substituting  this  already  modified  value  of  the 
original  volume,  V  (P/P'),  for  that  volume  denoted  by 
V  in  the  equation,  V  =  V  (T'/T),  representing  the  effect 
of  an  alteration  in  temperature.  In  other  words,  the  vol- 

21 


22  CHEMICAL   CALCULATIONS 

ume  V  in  the  equation  V'  =  V  (T'/T)  may  be  regarded  as 
already  affected  by  some  change  in  pressure,  —  a  change 
which  can  have  altered  it  only  in  accordance  with  the 
Law  of  Boyle  and  brought  it  to  its  new  value  through  the 
expression  V  (P/P').  It  follows,  therefore,  that  by  sub- 
stituting V  (P/P')  for  V  in  the  equation  V  =  V  (T'/T) 
we  shall  obtain  the  value  which  results  from  the  action  of 
both  these  agencies,  pressure  and  temperature,  upon  the 
gas.  Whether  we  consider  the  action  of  pressure  before 
or  after  that  of  temperature,  the  final  product  will  be  the 
same.  Thus,  when  the  expression  V  =  V  (T'/T)  is  made 
to  include  the  effect  of  a  change  in  pressure  upon  the 
original  volume  V,  we  have  the  form 

V  =  V(P/P')  (T'/T),or  V'  = 

whereas  the  expression  denoting  the  influence  of  pressure 
alone,  V'  =  V  (P/P'),  when  made  to  include  the  effect  of 
a  change  in  temperature  upon  the  volume  V,  gives  the 
same  form 

V  =  V  (T'/T)  <P/P'),  or  V  -  V     ,  . 


Though  the  Law  of  Boyle  holds  only  when  temperature 
is  a  constant  and  the  Law  of  Charles  only  when  pressure 
is  a  constant,  the  simultaneous  validity  of  the  two  laws  is 
here  assumed.  This  is  more  clearly  understood  when 
variations  in  temperature  and  pressure  are  studied  under 
constant  volume.  Thus  a  volume  of  gas  V,  at  tempera- 
ture T,  and  under  a  constant  pressure  P,  when  warmed  to 
the  temperature  T',  gives  a  new  volume  V,  according  to 
the  equation  V  =  V  (T'/T).  In  order  to  bring  back  this 
increased  volume  to  its  original  volume  V,  under  constant 
temperature,  a  pressure  P'  must  be  now  applied.  This 
pressure  will  be  derivable  from  the  original  pressure  in 
accordance  with  the  Law  of  Boyle;  and  from  the  new  vol- 


TEMPERATURE   AND   PRESSURE   UPON   GASES       23 

ume  V7,  or  V  (T'/T),  we  can  only  obtain  the  original  or 
smaller  volume  V  at  the  condition  of  this  greater  pressure 
P';  hence  the  equation  P'V  =  V  (T'/T)  P,  or  that  equa- 
tion which  denotes  the  inverse  proportionality  between 
volume  and  pressure.  With  the  volume  V,  thus  brought 
back  to  volume  V,  through  this  increase  in  pressure,  the 
variations  between  temperature  and  pressure  at  this  con- 
stant volume  can  be  derived.  The  elimination  of  the 
common  factor  V  from  the  equation  above  gives  us 

P'  =  p  (T'/T),  or  P'/P  =  T'/T. 

This  signifies  that  at  constant  volume  the  change  in 
pressure  is  directly  proportional  to  the  change  in  tem- 
perature expressed  upon  the  absolute  scale,  and  is  there- 
fore exactly  analogous  to  the  change  in  volume  by 
change  in  temperature  at  constant  pressure;  the  two 
thereby  stand  in  harmony  with  the  Law  of  Charles. 
The  influence  of  temperature  may  be  as  well  considered 
along  with  alterations  in  pressure  as  with  volume,  since 
the  alterations  are  exactly  analogous  and  are  derived 
from  identical  expressions.  Consequently  the  variations 
in  the  two  factors,  pressure  and  volume,  may  simulta- 
neously include  the  effect  of  temperature  and  take  that 
form  which  indicates  the  direct  proportionality  of  each 
of  these  factors  to  the  absolute  temperatures: 

T'/T  =  P'/P  X  V'/V. 

As  more  generally  expressed,  the  equation  is 
V'/V  =  P/P'  X  TVT,  or 


v  =  v  —  .  - 

\pr        'p 

This  is,  as  seen,  the  product  of  the  original  volume  by 
both  the  pressure-fraction  and  temperature-fraction  con- 
jointly. Hence  we  may  consider  the  original  volume  as 
having  come  under  the  action  of  two  forces  simultane- 


24  CHEMICAL   CALCULATIONS 

ously;  the  new  volume,  therefore,  is  that  volume  produced 
by  the  resultant  of  these  two  forces.* 

The  Relation  of  Volume  to  Temperature  and  Pressure. 

Example  8.  —  Given  100  c.c.  of  a  gas  at  10°  and  750  mm. 
pressure,  what  will  be  its  volume  at  standard  conditions  (0° 
and  760  mm.)? 

The  change  from  the  original  to  the  new  values  may  be 
indicated  thus:  283°  -*  273°,  and  750  mm.  ->  760  mm. 
With  respect  to  these  alterations  in  temperature  and  pres- 
sure the  questions  may  now  be  asked  in  regard  to  the 
magnitude  of  the  new  volume  brought  about  by  each  of 
these  influences.  From  the  change  in  temperature  the  new 
volume  must  be  smaller,  as  the  gas  in  cooling  undergoes 
a  contraction  (direct  proportionality),  hence  273/283  will 
represent  the  temperature-fraction.  From  the  change  in 
pressure  the  new  volume  must  be  smaller,  since  the  new 
pressure  is  larger  and  the  volume  therefore  will  be  dimin- 
ished (inverse  proportionality),  hence  750/760  will  repre- 
sent the  pressure-fraction.  Bringing  these  two  influences 
together,  the  product  of  the  two  by  the  original  volume 
will  give  the  combined  effect  of  the  two  changes  upon  this 
volume  : 

V7  =  100  (273/283)  X  (750/760),  or  V7  =  95.2  c.c. 

*  If  in  this  connection  the  form  of  equation  V  =  V  (1  +  at)  is  used, 
we  shall  arrive  at  a  similar  expression  for  change  in  pressure  at  con- 
stant volume  :P'  =  P  (1  +  at).  The  combined  effect  of  a  change  in 
volume  and  change  in  pressure  is  then  represented  by  P'V  =  PV 
(1  +  at).  By  substituting  here  for  t  its  value  on  the  absolute  scale, 
T  —  273,  we  obtain 

P'V  -  PV  P3  4-  T  "  21B\  or  P'V-  PV    T 
V+  '  ~'J 


273  273 

The  expression  PV/273  is  found  to  be  a  constant  for  a  given  quantity  of 
gas  under  all  values  of  temperature,  pressure  and  volume,  and  is  usu- 
ally expressed  by  K  ;  hence  the  general  gas  equation,  PV  =  RT,  which 
signifies  "that  the  pressure-volume  product  of  a  definite  quantity  of 
any  definite  gas  is  proportional  to  its  absolute  temperature."  — 
A.  A,  NOTES. 


TEMPERATURE   AND   PRESSURE    UPON    GASES        25 

With  these  two  fractions  each  less  than  unity  in  value, 
the  original  volume  is  materially  diminished  under  the 
new  conditions. 

The  Relation  of  Temperature  to  Volume  and  Pressure. 

Example  9.  —  At  what  temperature  will  a  gas  200  c.c.  in 
volume,  at  20°  and  750  mm.,  occupy  a  volume  of  300  c.c. 
when  the  pressure  is  740  mm.? 

Here  the  volumes  and  pressures  are  known,  and  the 
undetermined  quantity  is  one  of  temperature.  The 
general  formula  must  be  arranged  so  that  the  undeter- 
mined quantity  is  on  one  side  of  the  equation: 

V'/V  =  P/P'  X  T'/T 
when  divided  through  by  P/P'  gives 

T'/T  =  V'/V  X  P'/P, 
hence  the  new  temperature  T'  is  expressed  by  the  equation 


/V    P'\ 

Iv'p/' 


The  entire  matter  may  be  simplified  and  made  more 
apparent  if  we  recall  the  reasoning  advanced  in  the  pre- 
ceding cases;  namely,  the  unknown  quantity  is  derivable 
from  the  known  quantity  by  multiplying  this  latter  by 
the  fractions  formed  from  the  related  terms,  when  these 
terms  are  arranged  so  as  to  give  a  quotient  that  will 
express  the  magnitude  of  the  term  sought  (the  unknown 
quantity)  over  that  of  the  one  already  existent. 

At  20°  C.,  or  293°  A.,  the  volume  of  gas  in  the  problem 
above  was  200  .c.c.  At  the  new  temperature  it  is  to  be 
300  c.c.,  hence  this  temperature  must  be  larger,  else  no 
expansion  could  occur  (direct  proportionality) ;  therefore 
T'  =  293  (300/200).  Again,  the  pressure-fraction  must 
be  arranged  so  that  its  quotient  will  acquire  a  value 
less  than  unity,  i.e.,  740/750,  since  the  effect  of  a  lower 


26  CHEMICAL   CALCULATIONS 

pressure  must  necessarily  decrease  the  temperature  here 
required  (direct  proportionality)  (cf.  p.  23).  The  com- 
bined influence  will  be  expressed  by  the  product  of  the 
two  fractions  by  the  original  temperature: 

T'  =  293  (300/200)  X  (740/750) 
or  T'  =  433.6°  A.  =  160.6°  C. 

The  Relation  of  Pressure  to  Volume  and  Temperature. 

Example  10. —  100  c.c.  of  a  gas  at  20°  and  740  mm.  will 
exert  what  pressure  when  occupying  120  c.c.  at  40°? 

Arrange  the  related  terms  in  fractions  in  accordance 
with  the  questions  as  to  whether  the  new  pressure  is  to  be 
greater  or  less  than  the  original,  740  mm.,  and  multiply 
both  by  this  value.  The  temperature-fraction  becomes 
313/293,  since  at  higher  temperature  the  pressure  must 
be  greater  (direct  proportionality);  the  volume-fraction 
becomes  100/120,  since  at  increased  volume  the  pressure 
must  be  less  (inverse  proportionality),  therefore  by 
arranging  all  these  factors  we  have: 

P'  =  740  (100/120)  X  (313/293),  or  P'  =  658.7  mm. 

This  is  in  accord  with  the  general  equation  which  may  be 
drawn  up  for  changes  in  both  volume  and  temperature 
upon  the  pressure  of  a  gas: 


V_    T\ 
7"  T/ 


When  the  volume  is  a  constant  the  equation  takes  the 
form  indicated  on  p.  23: 

(rn/v 
1      \. 
I 

The  Relation  of  Density  to  Temperature  and  Pres- 
sure. —  The  variation  in  the  absolute  density  of  a  gas 
through  the  simultaneous  variations  in  temperature  and 


PARTIAL   PRESSURES  27 

pressure  follows  readily  from  what  has  been  stated  in 
regard  to  variation  in  volume  under  these  same  influences. 
In  the  study  of  absolute  density,  D  =  M/V,  the  inverse 
proportionality  between  the  absolute  density  and  the 
volume  of  a  gas  permits,  in  the  general  equation  denoting 
the  influence  of  temperature  and  pressure  changes  in  this 
connection,  of  the  substitution  of  the  term  D'  (the  new 
density)  and  D  (the  original  density)  for  V  and  V  re- 
spectively. The  general  equation  V'/V  =  P/P'  X  T'/T 
then  becomes  D/D'  =  P/P'  X  T'/T  or,  as  transposed  to 
indicate  the  alteration  in  temperature  and  pressure  upon 
the  original  density: 


This  is  in  keeping  with  what  has  been  already  observed, 
i.e.,  the  absolute  density  of  a  definite  quantity  of  gas 
varies  directly  with  the  pressure  and  indirectly  with  the 
absolute  temperature. 

Example  11.  —  A  volume  of  oxygen  at  20°  and  750  mm.  pres- 
sure has  what  absolute  density  under  these  conditions? 

The  absolute  density  of  oxygen  at  0°  and  760  mm. 
pressure  is  1.429.  The  new  density  will  be  less  under 
the  conditions  named,  since  both  the  pressure-fraction 
750/760,  and  temperature-fraction  273/293,  will  reduce 
the  original  value  according  to  the  equation: 

D'  =  1.429  (273/293)  (750/760),  or  D'  =  1.314. 
PARTIAL   PRESSURES. 

Whether  we  have  to  deal  with  a  simple  gas  or  a  mixture 
of  several  gaseous  components,  the  Law  of  Boyle  is  equally 
applicable.  Each  component  of  a  gaseous  mixture  exerts 
a  definite  individual  pressure,  the  same  that  it  would  exert 
were  it  alone  present  in  this  volume.  According  to  the 
Law  of  Dalton,  the  sum  of  these  individual  pressures, 


28  CHEMICAL   CALCULATIONS 

partial  pressures,  of  the  several  components  is  equal  to  the 
total  pressure  of  the  mixture.  When  two  gases  of  equal 
volume  at  the  same  conditions  are  brought  together  and 
communication  made  between  the  two  vessels,  we  find 
that  the  gas  from  each  vessel  diffuses  into  the  other  vessel 
and  becomes  uniformly  distributed  throughout  this  double 
volume  (the  two  vessels).  In  other  words,  each  gas  be- 
haves as  if  it  alone  were  present  in  the  total  space  included 
in  the  two  vessels.  Under  double  the  original  volume  each 
gas  can  exert  but  one-half  of  its  original  pressure.  The 
sum  of  the  pressures  of  the  two  gases,  however,  must  be 
again  equal  to  unity  or  that  pressure  which  each  originally 
exerted.  Consequently  the  total  pressure  of  a  gaseous 
mixture  may  be  regarded  as  the  sum  of  the  individual  or 
partial  pressures  of  the  several  components. 

When  aqueous  vapor  is  present  in  a  volume  of  gas  we 
have  again  the  consideration  of  gaseous  mixtures.  Water 
gives  off  a  varying  amount  of  its  vapor,  independent  of 
any  other  gaseous  substance  that  may  be  present  in  the 
space  about  it,  but  always  dependent  upon  the  tempera- 
ture. At  any  one  temperature  we  assume  an  equilibrium 
between  the  two  tendencies,  —  that  of  the  molecules  of 
the  vapor  to  condense  as  liquid,  and  that  of  the  molecules 
of  the  latter  to  fly  off  as  vapor;  in  other  words,  we  have  a 
condition  of  saturation  with  aqueous  vapor. 

The  density  or  concentration  of  aqueous  vapor  (the  aque- 
ous tension)  as  attained  at  each  condition  of  equilibrium 
is  definite  for  that  temperature  at  which  the  equilibrium 
exists.  This  density  of  the  vapor  is  determined  most  easily 
by  measuring  the  pressure,  which  it  exerts.  —  the  vapor 
pressure.  If  a  little  water  is  admitted  into  the  vacuum 
at  the  top  of  a  barometric  column  of  mercury,  the  vapor 
evolved  will  exert  a  pressure  which  increases  with  a  rise  in 
the  temperature  of  the  surrounding  medium.  At  100°  the 


PARTIAL   PRESSURES  29 

pressure  of  this  vapor  will  have  just  sufficed  to  displace  all 
of  the  mercury  (sustained  by  the  atmosphere  at  sea  level), 
and  consequently  will  have  reached  a  pressure  exactly  bal- 
ancing one  atmosphere.  This  temperature  is  the  boiling- 
point  of  the  liquid  or  that  point  at  which  bubbles  of  vapor 
form  in  the  liquid  itself.  At  higher  elevations  than  sea 
level,  i.e.,  under  reduced  atmospheric  pressures,  this 
attainment  of  the  external  pressure  by  the  vapor  of  the 
liquid  (the  boiling-point)  will  take  place  at  lower  tempera- 
tures. As  the  temperature  falls  the  aqueous  vapor  above 
the  mercury  in  the  barometric  tube  just  mentioned  will 
decrease  in  concentration  and  exert  less  pressure;  conse- 
quently mercury  will  be  forced  into  the  tube  by  the  pres- 
sure of  atmosphere  from  without.  The  difference  in  height 
of  the  mercury  in  this  tube  and  that  in  a  barometer,  where 
no  moisture  is  present,  will  give  the  height  of  mercury,  or 
pressure,  corresponding  to  that  of  the  aqueous  vapor  at 
the  recorded  temperature. 

When  any  gas  is  measured  over  water  at  atmospheric 
pressure,  and  a  temperature  constant  for  both  gas  and 
liquid,  we  obtain  the  partial  pressure  of  the  dry  gas  in  the 
mixture  by  subtracting  from  this  barometric  reading  the 
pressure  which  aqueous  vapor  exerts  at  the  observed  tem- 
perature. (The  values  for  a  range  of  temperatures  are 
found  in  Appendix  II.) 

Example  12.  — 100  c.c.  of  a  gas  at  10°  and  750  mm.,  measured 
over  water,  will  have  what  volume  under  standard  conditions? 

At  10°  the  pressure  of  aqueous  vapor  is  9.2  mm.  (see 
Appendix  II),  therefore  the  barometric  reading,  750  mm., 
represents  the  sum  of  the  pressures  of  the  aqueous  vapor 
and  that  of  the  dry  gas,  and,  as  standard  conditions  are 
desired,  the  volume  of  the  gas  should  be  determined  in  the 
dry  state.  Accordingly  750  mm.  —  9.2  mm.  =  740.8  mm., 
the  pressure  of  the  dry  gas.  The  problem  then  resolve? 


30  CHEMICAL   CALCULATIONS 

itself  into  one  exactly  as  in  Ex.  8,  page  24.  The  volume 
of  a  gas  at  740.8  mm.,  when  calculated  at  a  pressure  of 
760  mm.;  or  a  greater  pressure,  must  be  necessarily 
diminished;  therefore  740.8/760  is  the  pressure-fraction. 
The  temperature-fraction  is  273/283,  consequently  the 
new  volume  will  be  derived  from  the  equation 

V  =  100  X  740.8/760  X  273/283,  or  V  =  94.03  c.c. 

PROBLEMS. 

18.  500  c.c.  of  hydrogen  at  25°  and  745  mm.  pressure  will  have 
what  volume  at  15°  and  755  mm.  pressure?     Ans.  476.8  c.c. 

19.  A  vessel  of  2000  c.c.  capacity  held  5  grams  of  a  vapor 
at  the  standard  conditions  of  temperature  and  pressure.     What 
weight  of  this  vapor  at  10°  and  750  mm.  pressure  can  be  held  in 
this  vessel,  the  capacity  considered  constant? 

Ans.  4.76  grams. 

Note.  — This  volume  of  vapor  (x),  at  0°  and  760  mm.,  with  the 
weight  (w>),  when  raised  to  the  new  volume  (?/),  will  still  have  the  same 
weight.  Consequently  the  vessel  can  be  made  to  contain  only  that 
part  of  the  original  weight  which  is  denoted  by  the  fraction  x/y. 

20.  10  liters  of  a  gas  measured  at  20°  and  750  mm.  pressure 
weighed  14  grams.     What  weight  of  this  gas  could  be  contained 
in  a  smaller  vessel  holding  4  liters  at  10°  and  760  mm.  pressure? 

Ans.  5.87  grams. 

21.  What  volume  will  1000  c.c.  of  gas  at  30°  and  740  mm. 
pressure  occupy  when  reduced  to  standard  conditions? 

Ans.  877.3  c.c. 

22.  1000  c.c.  of  a  gas  measured  at  10°  and  750  mm.  pressure 
was  increased  to  1120  c.c.  by  warming.     The  final  pressure  read 
740  mm.     What  was  the  final  temperature  of  the  gas? 

Ans.    39.7°. 

23.  A  volume  of  gas  measured  at  10°  and  750  mm.  pres- 
sure will  have  what  pressure  at  20°,  the  volume  a  constant? 

Ans.  776.5  mm. 

24.  A   volume    of   gas   measuring    1000   c.c.    at   15°   and 
745  mm.  pressure  was  warmed  to  32°.     What  increase  in  the 
pressure  of  the  gas  would  be  recorded,  the  volume  a  constant? 

Ans.  44  mm. 


TEMPERATURE  AND  PRESSURE  UPON  GASES   31 

25.  What  decrease  in  temperature  will  be  necessary  to  reduce 
400  c.c.  of  a  gas,  at  20°  and  765  mm.,  to  a  volume  of  300  c.c.  at 
750mm.?  Ans.    77.56°. 

26.  What  increase  in  temperature  will  be  necessary  to  bring 
a  volume  of  gas,  measuring  560  c.c.  at  10°  and  745  mm.  pres- 
sure, to  a  volume  of  600  c.c.  at  this  same  pressure? 

Ans.    20.2°. 

27.  A  volume  of  hydrogen  measuring  500  c.c.  at  25°  and 
730  mm.  was  reduced  in  volume  to  400  c.c.  at  0°.     What  was 
the  final  pressure  of  the  gas?  Ans.    836  mm. 

28.  What  increase  in  pressure  is  necessary  to  force  100  c.c.  of 
hydrogen,  at  10°  and  740  mm.,  into  a  vessel  of  80  c.c.  capacity, 
when  the  temperature  of  this  vessel  is  constant  at  0°? 

Ans.    152.3  mm. 

29.  What  is  the  absolute  density  of  hydrogen  at  20°  and 
740  mm.  pressure?  Ans.    0.08153. 

30.  What  is  the  absolute  density  of  air  at  10°  and  750  mm. 
pressure?    The  absolute  density  at  the  standard  conditions  is 
1.293.  Ans.    1.231. 

31.  400  c.c.  of  gas,  with  the  absolute  density  6  and  measured 
at  25°  and  750  mm.  pressure,  is  to  be  brought  to  a  temperature 
of  10°  and  a  pressure  of  760  mm.     What  weight  of  this  final  gas 
can  be  contained  in  a  vessel  of  100  c.c.  capacity? 

Ans.    0.64  gram. 

32.  10  grams  of  a  gas,  measured  at  -  48°  and  600  mm.  pres- 
sure, was  expanded  by  heating  to  177°  and  reducing  the  pressure 
to  480  mm.     Of  this  rarefied  gas  250  c.c.  weighed  0.5  gram. 
What  was  the  original  volume  of  gas  and  what  was  the  density  of 
the  gas  at  its  original  and  final  volume? 

Ans.  2000  c.c. 
Original  density  =  5. 
Final  density      =  2. 

33.  150  c.c.  of  air  measured  over  water  at  18°  and  746  mm. 
pressure  will  have  what  volume  at  standard  conditions? 

Ans.    135.3  c.c. 

34.  1  liter  of  oxygen  at  standard  conditions  weighs  1.429 
grams.    440  c.c.  of  this  gas  measured  over  water  at  24°  and 
742  mm.  pressure  will  contain  what  weight  of  the  dry  gas? 

Ans.    0.547  gram. 


32  CHEMICAL   CALCULATIONS 

35.  545  c.c.  of  nitrogen  as  measured  over  water  at  22°  and 
748  mm.  pressure  contain  what  weight  of  the  dry  gas  (density  = 
1.2507)?  Ans.    0.6045  gram. 

36.  2.2  grams  of  oxygen  will  occupy  what  volume  at  20°  and 
770  mm.  pressure  when  transferred  to  a  vessel  inverted  over 
water?  Ans.    1667  c.c. 

37.  1  gram  of  hydrogen  (density  =  0.08987)  measured  at 
standard  conditions  will  occupy  what  volume  when  transferred 
to  a  vessel  over  water  at  0°  and  760  mm.  pressure? 

Ans.  11,195  c.c. 

38.  400  c.c.  of  oxygen  at  standard  conditions  will  have  what 
volume  when  measured  over  water  at  20°  and  755  mm.  pressure? 

Ans.    442.3  c.c. 

39.  100  c.c.  of  a  gas,  measured  over  water  at  25°  and  745  mm. 
pressure,  will  have  what  volume  when  deprived  of  moisture? 
The  atmospheric  conditions  constant.  Ans.  96.8  c.c. 

40.  1000  c.c.  of  oxygen,  measured  over  water  at  10°  and 
750  mm.  pressure,  will  have  what  volume  at  -  10°  and  770  mm. 
pressure  when  deprived  of  its  moisture?         Ans.  894.1  c.c. 

41.  500  c.c.  of  a  gas  contained  in  a  tube  inverted  over 
water,  and  measured  at  10°  and  765  mm.  pressure,  will  have 
what  volume  under  a  change  in  the  atmospheric  conditions  to 
20°  and  745  him.  pressure?  Ans.   537.7  c.c. 

42.  What  decrease  in  pressure  will  be  necessary  to  raise  a 
volume  of   gas  measuring  over  water  400  c.c.,  at   22.5°   and 
748  mm.  pressure,  to  a  volume  of  440  c.c.  under  the   same 
conditions?  Ans.   66.2  mm. 

Note.  — The  actual  atmospheric  pressure  of  the  moist  gas  required 
for  the  new  conditions  will  be  obtained  by  adding  the  tension  of  aque- 
ous vapor  to  the  value  for  P'. 

43.  What  increase  in  atmospheric  pressure  will  be  necessary 
to  reduce  200  c.c.  of  a  gas,  measured  in  a  tube  over  water  at  10° 
and  720  mm.,  to  a  volume  of  100  c.c.  at  20°  in  this  same  tube? 

Ans.   769.2  mm. 

Note.  — The  calculated  pressure  for  the  dry  gas  must  be  increased 
by  the  tension  of  aqueous  rapor  at  the  new  conditions  in  order  to 
obtain  the  actual  pressure  that  would  be  recorded  (cf.  Prob.  42). 


CHAPTER  VI. 

AVOGADRO'S   HYPOTHESIS   AND   SOME   OP   ITS 
APPLICATIONS. 

THE  hypothesis  of  Avogadro  is  the  outcome  of  the  Law 
of  Combining  Volumes  (Gay-Lussac),  and  assumes  that,  if 
the  Laws  of  Boyle  and  Charles  are  strictly  true,  there 
must  be  a  uniform  distribution  of  molecules  in  all  volumes 
of  gases  at  the  same  conditions  of  temperature  and  pres- 
sure. More  generally  the  hypothesis  takes  the  following 
form:  Under  the  same  conditions  of  temperature  and 
pressure  equal  volumes  of  all  substances  in  the  state  of 
vapor  contain  an  equal  number  of  molecules.  By  mole- 
cules are  meant  the  smallest  parts  in  whicR  a  substance 
maintains  its  identity.  The  complete  chemical  combina- 
tion between  the  molecules  of  equal  or  multiple  volumes 
of  gases  to  form  definitely  related  volumes  of  gaseous 
products,  with  no  molecules  of  either  constituent  remain- 
ing free  or  uncombined,  presupposes  this  even  distribution  t 
of  molecules. 

Now  the  weight  of  a  given  volume  of  gas  when  compared 
with  the  weight  of  an  equal  volume  of  another  gas,  under 
the  same  conditions,  will  represent  not,  alone  the  ratio 
between  the  sum  of  the  weights  of  all  the  molecules  in  one 
volume  and  that  of  the  other,  but  also  the  ratio  between 
the  weight  of  an  individual  molecule  of  one  gas  and  that 
of  the  other.  This  follows,  of  course,  from  the  fact  that 
the  numerical  ratio  between  the  weights  of  each  equaf 
volume  is  not  altered  through  division  by  a  common  factor, 
—  the  unknown  yet  equal  number  of  molecules  in  both 

33 


34  CHEMICAL   CALCULATIONS 

volumes.  In  determining  the  relative  weights  of  mole- 
cules it  is  therefore  of  no  importance  as  to  what  particular 
volume  of  gas  is  weighed,  so  long  as  the  weight  is  to  be 
compared  with  the  weight  of  an  exactly  equal  volume  of 
some  other  gas  at  the  same  conditions  of  temperature  and 
pressure.  Any  known  volumes  of  gases  upon  which  accu- 
rate data  are  at  hand  may  serve  for  this  purpose  when 
they  are  reduced  to  the  same  conditions  (cf.  Chap.  V)  and 
comparison  made  between  equal  volumes  of  each. 

The  Relation  of  Density  to  Molecular  Weight.  —  This 
comparison  in  the  weights  of  equal  volumes  of  gases  is, 
after  all,  nothing  more  or  less  than  the  comparison  in  gas 
densities.  Thus  the  absolute  density  of  hydrogen,  0.08987, 
when  compared  with  that  of  oxygen,  1.429,  is  found  to  be 
about  1/16  of  the  latter,  or,  if  the  density  of  hydrogen 
is  considered  as  unity,  the  density  of  oxygen  becomes 
15.90: 

0.08987  :  1.429  =  1  :  15.90. 

The  relative  weight  of  a  molecule  of  oxygen,  —  its  molecu- 
lar weight,  —  would  be,  accordingly,  15.90  times  the 
molecular  weight  of  hydrogen.  But  from  accurate  deter- 
minations of  the  atomic  weights  of  these  elements  the 
molecular  weight  of  oxygen  is  found  to  be  only  15.88 
times  that  of  hydrogen.* 

*  The  determination  of  the  atomic  weights  of  all  the  elements  gives 
us  the  most  accurate  means  of  obtaining  the  true  molecular  weights. 
Density  determinations  involve  numerous  complications,  and  are  there- 
fore liable  to  errors,  but  nevertheless  they  give  us  a  very  fair  approach 
to  the  true  values.  The  various  degrees  of  cohesion  between  the 
molecules  of  gases  bring  into  existence  a  greater  or  less  deviation  from 
the  uniform  packing  of  these  molecules  as  assumed  by  Avogadro's 
hypothesis  for  a  perfect  gas.  Thus  with  oxygen  we  have  a  slightly 
greater  packing  than  with  hydrogen,  as  is  shown  by  the  increase  of 
the  relative  density  15.90  over  the  calculated  value  15.88.  In  order 
that  our  calculations  may  be  free  of  these  slight  discrepancies,  densi- 
ties may  be  given  as  calculated  back  from  the  correct  molecular  weights, 


AVOGADRO'S    HYPOTHESIS  35 

The  relation  in  the  absolute  densities  applies  equally 
well  to  the  relative  densities  of  gases  providing,  of  course, 
that  they  are  referred  to  the  same  standard.  Thus  the 
relative  densities  of  hydrogen  and  oxygen  upon  air  as  a 
standard  are  0.0696  and  1.105  respectively.  The  com- 
parison of  these  values  gives  the  same  relative  weights 
of  the  molecules  of  hydrogen  and  oxygen  as  shown 
above. 

We  have  several  cases  which  show  that  a  volume  of 
hydrogen  may  enter  into  combination  with  an  exactly 
equal  volume  of  another  gas,  and  the  resulting  gaseous 
compound  occupy  the  volume  originally  held  by  the  two 
gases  severally,  i.e.,  the  sum  of  the  two  gaseous  volumes 
now  makes  up  the  volume  of  the  compound.  Since 
temperature  and  pressure  remain  the  same  throughout, 
the  number  of  molecules  in  the  volume  of  the  resulting 
compound  must  be  twice  the  number  of  molecules  in 
either  volume  of  gaseous  constituent.  As  each  of  the 
molecules  in  the  compound  contains  hydrogen  in  chemical 
combination,  we  clearly  see  that  there  are  now  twice  as 
many  hydrogen  parts  present  as  there  were  in  the  original 
volume  of  hydrogen.  Since  weight  for  weight  there  can 
be  no  change  in  the  amount  of  hydrogen  present  in  the 
two  cases,  we  naturally  infer  that  the  original  molecule  of 
hydrogen  contained  two  smaller  parts  or  units,  —  atoms,  — 
and  that  it  was  these  atoms  which  were  concerned  in 
the  chemical  combination.  Never  have  we  been  able  to 
get  the  element  hydrogen  to  spread  out  over  more  than 
twice  its  original  volume  when  entering  into  chemical 
combination,  therefore  we  assume  that  the  hydrogen  as 
we  know  it  contains  at  least  two  atoms  to  its  molecule. 

and  consequently  the  value  for  oxygen  will  henceforth  be  taken  as 
15.88  instead  of  15.90,  when  referred  to  hydrogen  as  unity. 

Unless  otherwise  stated,  the  densities  hereafter  considered  are  the 
calculated  weights  of  1  liter  of  gas  at  standard  conditions. 


36  CHEMICAL    CALCULATIONS 

We  have  a  number  of  elements  the  molecules  of  which 
under  certain  conditions  are  shown  in  like  manner  to 
possess  two  atoms  (i.e.,  diatomic),  e.g.,  oxygen  (02), 
chlorine  (C12),  bromine  (Br2),  iodine  (I2),  and  nitrogen  (N2), 
whereas  the  molecule  of  mercury  in  the  vapor  state  is 
monatomic  (Hg). 

As  already  stated,  from  the  comparison  in  the  weights 
of  equal  volumes  of  gases,  we  may  derive  the  relative 
weights  of  the  several  kinds  of  molecules.  With  the 
hypothesis  now  that  the  molecules  themselves  are  com- 
posed of  atoms,  we  must  interpret  these  relative  values  as 
those  which  stand  for  the  ratios  between  the  sums  of  the 
weights  of  the  atoms  in  the  molecules,  i.e.,  the  molecular 
weights.  A  comparison  between  the  relative  weight  of  a 
molecule  of  hydrogen  and  a  molecule  of  oxygen  gives  us 
the  ratio  1  :  15.88.  Since  both  hydrogen  and  oxygen 
molecules  contain  two  atoms  each,  this  same  ratio  stands 
likewise  for  that  between  the  weights  of  the  hydrogen  and 
oxygen  atoms.  In  order,  therefore,  to  include  the  relative 
weights  of  the  atoms  in  the  molecular  weights  and  thus  be 
able  to  compare  the  relative  weights  of  the  molecules  upon 
their  smallest  units  as  a  basis,  it  is  only  natural  that  we 
regard  the  ratio  1  :  15.88  as  the  ratio  between  the  relative 
weights  of  the  hydrogen  and  oxygen  atoms.  The  molecu- 
lar, weights  of  hydrogen  and  oxygen  then  become  2  and 
31.76  respectively,  values  which  indicate,  of  course,  that 
they  are  made  up  of  the  sums  of  the  respective  atomic 
weights  in  each  molecule.  All  of  this  presupposes  that 
hydrogen,  with  the  smallest  atomic  weight,  should  be  re- 
garded as  unity  for  the  convenience  of  these  comparisons. 

The  importance  of  the  element  oxygen  with  its  great- 
number  of  well-defined  compounds  and  the  comparative 
ease  with  which  these  compounds  may  be  utilized,  through 
analyses,  for  the  determination  of  the  relative  weights  of 


AVOGADRO'S    HYPOTHESIS  37 

molecules  of  other  elements,  has  led  to  its  universal  adop- 
tion as  the  standard  of  molecular  weights.  In  order  that 
the  values  upon  this  scale  may  not  depart  far  from  those 
already  determined  upon  the  basis  of  hydrogen  as  unity, 
the  former  value  for  oxygen,  31.76,  was  increased  to  32,  a 
whole  number,  and  thus  the  atomic  weights  of  all  the 
elements  allowed  to  retain  a  value  greater  than  unity. 
With  oxygen  changed  from  31.76  to  32,  we  may  readily 
derive  the  corresponding  value  for  the  molecular  weight 
of  hydrogen  from  the  simple  proportion: 

31.76  :  2  =  32  :  x, 

which  gives  the  value  2.016.  From  the  molecular  weight 
of  oxygen  (32)  the  relative  weight  of  the  atom  must  be- 
come 16,  while  the  relative  weight  of  the  hydrogen  atom 
will  be  1.008.  An  imaginary  gas  with  the  atomic  weight 
1/16  of  the  atomic  weight  of  oxygen  (16)  may  be  regarded, 
therefore,  as  the  basis  or  unit  weight  of  these  values. 
These  numerical  values  refer  to  no  particular  standard  of 
weights,  but  when  expressed  in  grams  it  is  customary  to 
call  them  gram-molecular  weights;  thus  32  grams  is  the 
G.M.W.  of  oxygen. 

As  already  stated,  the  comparison  of  the  densities  of  all 
substances  in  the  state  of  vapor  gives  at  once  the  relative 
weights  of  their  molecules.  In  order,  then,  to  obtain  the 
molecular  weight  of  any  substance  upon  the  basis  of  oxy- 
gen as  32  it  becomes  necessary  to  effect  a  comparison 
between  its  absolute  density  and  that  of  oxygen  on  this 
basis.  The  calculated  absolute  density  of  chlorine  is 
3.166  (actual  value  =  3.22).  The  absolute  density  of 
oxygen,  1.429,  will  bear  the  same  relation  to  this  value 
for  chlorine  as  the  weight  32  bears  to  the  molecular  weight 
of  chlorine,  70.92: 

1.429  :  3.166  =  32  :  x,  or  x  =  70.92. 


38  CHEMICAL   CALCULATIONS 

When  the  relative  density  of  a  gaseous  substance  is 
given,  i.e.,  the  value  obtained  by  referring  the  absolute 
density  to  that  of  oxygen  as  unity,  the  proportion  above 
will  of  course  take  the  form: 

1  :  Rel.  density  of  gas  =  32  :  x. 

This  proportion  shows  that  the  molecular  weight  (x)  is 
merely  the  product  of  the  relative  density  by  32.  This 
is  the  logical  outcome  of  making  oxygen  the  standard 
both  of  gas  densities  and  molecular  weights.  Since  as  the 
basis  of  relative  densities  the  weight  of  a  unit  volume  of 
oxygen  is  considered  unity  and  as  the  basis  of  molecular 
weights  it  is  considered  32,  we  naturally  need  only  multiply 
any  particular  relative  density  by  32  in  order  to  obtain 
the  molecular  weight  of  this  substance. 

If,  on  the  other  hand,  the  relative  density  of  a  gas  is 
given  in  terms  of  some  other  density  besides  oxygen  as  a 
standard,  the  molecular  weight  of  this  standard  substance 
must  of  course  replace  32  in  the  proportion  above.  For- 
merly the  relative  densities  were  determined  upon  the 
basis  of  hydrogen  as  unity.  The  proportion  indicating 
the  molecular  weight  upon  this  basis  is  as  follows: 

1  :  Rel.  density  (H  =  l)  =  2.016  :  x. 

Consequently  the  molecular  weight  may  be  derived  from 
the  relative  density  (H  =  l)  by  multiplying  this  value 
by  2.016. 

The  Calculation  of  Relative  Densities  upon  Different 
Standards.  —  To  convert  the  relative  density  of  a  gas 
upon  the  hydrogen  standard  over  to  the  standard  of  oxy- 
gen, or  vice  versa,  we  shall  need  to  refer  this  value  to  the 
ratio  between  the  absolute  densities  of  the  two  standards, 
namely  0.08987  and  1.429,  in  an  inverse  order.  As  the 
means  and  extremes  of  the  proportion  we  must  have 
the  product  of  the  relative  density  of  a  substance  by  the 


AVOGADRO'S   HYPOTHESIS 


39 


absolute  density  of  the  corresponding  standard,  a  product 
always  equal  to  the  absolute  density  of  the  substance  in 
question;  and  hence  a  constant  for  the  proportion.  Thus: 


Abs.  density 
oxygen 
(1.429) 


Abs.  density 
hydrogen 
(0.08987) 


Rel.  density 

substance 

(H-l) 


Rel.  density 

substance 

(0  =  1). 


Example  13.  —  The  calculated  relative  density  of  chlorine 
(H  =  1)  is  35.23.  What  is  the  relative  density  of  this  gas 
referred  to  oxygen? 

According  to  the  proportion:  1.429:  0.08987  =  35.23  :  x, 
the  value  of  x,  or  relative  density,  (0  =  1),  is  calculated 
as  2.216.  When  this  is  multiplied  by  32  we  obtain  the 
molecular  weight  of  chlorine,  70.92. 

Quite  often  the  vapor  density  of  a  substance  is  deter- 
mined upon  air  as  a  standard.  The  ratio  1.429  :  1.293  is 
the  ratio  between  the  absolute  densities  of  oxygen  and  air 
respectively,  hence  the  calculated  molecular  weight  of  air 
referred  to  oxygen  as  32  will  be  derived  from  the  propor- 
tion: 1.429  :  1.293  =  32  :  x.  From  this  the  value  28.955 
is  found;  a  value  which  signifies  that  if  air  were  a  com- 
pound its  molecular  weight  would  be  28.955. 

The  molecular  weight  of  a  substance  may  be  calculated, 
therefore,  from  its  vapor  density  referred  to  air  if  we 
substitute  this  value,  28.955,  for  32  in  the  proportion  on 
page  38,  and  the  relative  vapor  densities  are  made  one 
member  of  the  proportion. 

Example  14.  —  The  calculated  vapor  density  of  mercury  is 
6.908  referred  to  air.  What  is  its  molecular  weight? 

The  proportion  1  :  6.908  =  28.955  :  x  gives  200  as  the 
molecular  weight  of  mercury.  The  conversion  of  any 
vapor  density  (air  =  1)  to  the  .basis  (0  =  1)  is  exactly 
analogous  to  the  case  of  hydrogen  and  oxygen  above. 
The  absolute  density  of  air  is  1.293. 


40  CHEMICAL   CALCULATIONS 

The  Gram-Molecular  Volume.  —  If  the  actual  weight  of 
1  c.c.  of  a  gas  is  known  in  grams  and  this  value  is  divided 
into  the  molecular  weight  also  expressed  in  grams,  the 
quotient  will  indicate  the  exact  number  of  cubic  centi- 
meters of  the  gas  in  question  that  will  be  required  to  give 
this  weight  or  the  gram-molecular  weight  (G.M.W.).  In 
other  words,  the  quotient  expresses  the  volume  in  cubic 
centimeters  which  contains  the  gram-molecular  weight. 

For  example,  32  =  G.M.W.  of  oxygen;  0.001429  gram 
is  the  weight  of  1  c.c.  of  oxygen,  therefore 

32 


0.001429 


=  22,390  (approximately), 


the  number  of  cubic  centimeters  of  oxygen,  at  0°  and 
760  mm.,  necessary  to  give  its  G.M.W. 

Again,  the  G.M.W.  of  hydrogen  is  2.016  and  the  weight 
of  1  c.c.  is  0.00008987  gram,  therefore 

22,400  (approximately), 


0.00008987 

the  number  of  cubic  centimeters  of  hydrogen  at  0°  and 
760  mm.  necessary  to  give  its  G.M.W. 

This  volume  occupied  by  the  G.M.W.  of  a  substance 
is  known  as  the  gram-molecular  volume  (G.M.V.).  Its 
calculation  from  the  density  of  a  gas  gives  values  which 
diverge  slightly  from  the  average  obtained  for  the  more 
nearly  perfect  gases  (i.e.,  22,400  c.c.)  according  as  the 
degree  of  packing  of  the  molecules  in  each  diverges  from 
that  in  the  latter.  We  deduce,  therefore,  the  simple 
statement  that  the  weight  of  22,400  c.c.  of  a  gaseous  sub- 
stance, at  0°  and  760  mm.,  will  give,  when  expressed  in 
grams,  the  gram-molecular  weight  of  that  substance. 
From  the  weight  of  any  given  volume  of  a  gas  the  weight 
of  22,400  c.c.  may  be  calculated  by  simple  proportion; 


AVOGADRO'S   HYPOTHESIS  41 

hence  a  ready  means  is  given  for  determining  molecular 
weights  in  general. 

Example  15.  —  What  is  the  molecular  weight  of  a  gas,  5600  c.c. 
of  which  at  standard  conditions  weigh  5  grams? 

The  value  22,400  is  a  standard,  and  the  volume  here  to 
be  compared  with  it  (5600)  must  form  the  same  ratio 
thereto  as  the  relative  weights  of  the  two  volumes: 

5600  :  22,400  =  5  :  x,  or 

From  which  the  molecular  weight  of  this  gas  (x)  is  found 
to  be  20. 

With  this  definite  relation  established  between  the 
volume  22,400  c.c.,  calculated  at  0°  and  760  mm.,  and  that 
weight  in  grams  which  stands  for  the  molecular  weight 
of  an  element  or  compound,  we  may  derive  the  fractional 
volume  of  this  G.M.V.  which  any  corresponding  fractional 
part  of  the  G.M.W.  of  a  substance  may  occupy,  and,  vice 
versa,  any  fractional  weight  of  the  G.M.W.  which  any 
fractional  part  of  this  G.M.V.  of  the  substance  in  state  of 
vapor  may  have.  In  other  words,  the  same  fractional 
parts  of  the  G.M.V.  and  the  G.M.W.  are  directly  pro- 
portional to  each  other. 

Example  16.  —  16  grams  of  oxygen  will  occupy  what  vol- 
ume at  0°  and  760  mm.  pressure? 

Since  32  grams  occupy  22,400  c.c.  at  these  conditions, 
the  volume  occupied  by  16  grams  will  be  16/32  of  22,400 
or  11,200  c.c.,or,  by  simple  proportion,  32  : 16  =  22,400  :  x, 
or  x  =  11,200  c.c. 

To  apply  the  effect  of  temperature  and  pressure  changes 
let  it  be  desired  to  find  the  volume  which  2  grams  of 
nitrogen  will  occupy  at  20°  and  750  mm.  The  molecular 
weight  of  nitrogen  is  28.  22,400  :  x  =  28: 2,  or  x  =  1600  c.c. 
This  volume  corrected  for  temperature  and  pressure  be- 


42  CHEMICAL   CALCULATIONS 

comes  increased  at  20°  by  293/273  of  itself,  and,  at  750  mm., 
by  760/750  of  itself,  and  hence  the  corrected  volume  will 


be  1600  (ff  Xj-f),  or  1739.2  c.c. 


Vice  versa,  the  weight  of  any  volume  of  a  substance  in 
the  state  of  vapor  may  be  determined  if  the  gram-molecu- 
lar weight  of  the  compound  is  known. 

Example  17.  —  What  is  the  weight  of  5600  c.c.  of  hydrogen 
chloride  measured  at  0°  and  760  mm.? 

The  molecular  weight  of  hydrogen  chloride  is  35.46  + 
1.01,  or  36.47.  Therefore 

22,400  :  5600  =  36.47  :  x,  or  x  =  9.118  grams. 

Calculation  of  Densities  of  Gases  from  the  Molecular 
Weights.  —  From  these  relations  the  determination  of  the 
density  of  any  volume  of  vapor  is  reduced  to  the  simple 
task  of  ascertaining  from  the  molecular  weight  of  the 
substance  and  this  volume  22,400  c.c.,  what  the  weight 
of  a  unit  volume  (1  liter  for  gases)  will  be  at  standard  con- 
ditions. The  determination  of  the  relative  density  of  one 
gaseous  substance  upon  any  other  as  a  standard  resolves 
itself,  therefore,  into  a  comparison  of  the  gram-molecular 
weights  of  the  two  gases,  i.e.,  the  comparison  of  the 
weights  of  these  equal  volumes,  —  the  G.M.V.  With  air, 
the  weight  of  22,400  c.c.  has  been  given  as  28.955  grams 
(this  may  be  construed  as  the  hypothetical  gram-molecular 
weight  of  air). 

Example  18.  —  What  is  the  relative  density  of  chlorine  re- 
ferred to  (a)  air?  to  (6)  oxygen? 

(a)  The  gram-molecular  weight  of  chlorine  is  70.92,  the 
weight  of  22,400  c.c.  The  weight  of  22,400  c.c.  of  air  is 
28.955  grams.  Therefore  the  ratio  70.92/28.955  when  made 
equal  to  the  ratio  x/l  (where  the  weight  of  an  equal  volume 


AVOGADKO'S  HYPOTHESIS  43 

of  air  is  unity)  gives  us  the  proportion :  70.92:  28.955  =  x  :  1, 
or  the  relative  density  of  chlorine  as  2.449  (air  =1). 

(6)  When  referred  to  oxygen  the  relative  density  of 
chlorine  is  determined  in  an  exactly  similar  manner.  The 
G.M.W.  of  oxygen,  32,  replaces  the  weight  of  22,400  c.c. 
of  air  in  the  example  above.  The  process  is  of  course  the 
reverse  of  that  in  which  the  molecular  weight  of  a  sub- 
stance is  derived  by  multiplying  its  relative  density  (O  =  1 ) 
by  the  value  32,  the  basis  of  molecular  weights  (O  =  32) 
(cf.  p.  38). 

PROBLEMS. 

44.  The  relative  density  of  carbon  dioxide  (H=l)  is  21.83. 
What  is  the  relative  density  of  this  gas  upon  the  oxygen  stand- 
ard? Ans.  1.373. 

45.  The  relative  density  of  chlorine  (air  =  1)  is  2.449.    What 
is  its  relative  density  upon  the  oxygen  standard? 

Ans.  2.216. 

46.  The  vapor  density  of  phosphorus  trichloride  (air  =  l) 
is  4.745.     What  is  its  molecular  weight?  Ans.  137.38. 

Note.  — Though  Ex.  14  may  be  followed  it  is  much  more  simple 
to  calculate  the  weight  of  1  liter  of  the  vapor,  which  is  4.745  times  the 
weight  of  1  liter  of  air  (1.293  grams),  and  with  this  weight  of  1  liter 
to  apply  Ex.  15. 

47.  What  is  the  molecular  weight  of  mercuric  chloride,  the 
vapor  density  of  which  is  9.354  referred  to  air? 

Ans.  270.84. 

48.  The  vapor  density  of  water  is  0.622   (air  =  l).     What 
is  its  molecular  weight?  Ans.  18.016. 

49.  What  is  the  molecular  weight  of  sulphur  dioxide,  the 
relative  density  of  which  is  2.002  (O  =  l)?  Ans.  64.07. 

50.  What  is  the  molecular  weight  of  a  gas,  2  liters  of  which  at 
standard  conditions  weigh  12  grams?  Ans.  134.4. 

51.  The  weight  of  3840  c.c.  of  a  certain  vapor,  at  standard 
conditions,  is  24  grams.     What  is  the  molecular  weight  of  the 
substance?  Ans.  140. 


44  CHEMICAL   CALCULATIONS 

52.  3180  c.c.  of  a  gas  measured  at  24°  and  750.2  mm.  pressure 
weighed  6  grams.     What  is  the  molecular  weight?     Ans.   48. 

53.  8019  c.c.  of  a  gas  measured  over  water  at  20°  and 
742.4  mm.  pressure  weighed  14  grams  when  deprived  of  aque- 
ous vapor.     Calculate  the  molecular  weight.  Ans.  44. 

54.  5647  c.c.  of  a  gas  measured   over  water  at  24°  and 
754.2  mm.  pressure  weighed  6.254  grams  when  deprived  of  aque- 
ous vapor.     Calculate  its  molecular  weight.  Ans.  28.02. 

55.  What  volume  will  49.63  grams  of  chlorine  (mol.  wt.  = 
70.92)  occupy  at  standard  conditions?  Ans.  15,680  c.c. 

56.  What  is  the  volume  occupied  by  8.8  grams  of  carbon 
dioxide  (mol  wt.  =  44)  at  12°  and  752  mm.  pressure? 

Ans.  4726  c.c. 

57.  10  grams  of  carbon  dioxide  (mol.  wt.  =  44)  will  occupy 
what  volume  when  contained  in  a  vessel  over  water  at  20°  and 
742.4  mm.  pressure?  Ans.  5728  c.c. 

58.  1  gram  of  oxygen  will  occupy  what  volume  over  water 
at  30°  and  756.5  mm.  pressure?  Ans.  814.3  c.c. 

59.  8  grams  of  hydrogen  (mol.  wt.  =  2.016)  are  to  be  ad- 
mitted into  a  balloon  at  a  temperature  of  20°  and  a  pressure  of 
740  mm.    What  must  be  the  capacity  of  the  balloon? 

Ans.  97,980  c.c. 

60.  Of  what  capacity  is  that  vessel  which  contains  4  grams 
of  oxygen  (mol.  wt.  =  32)  at  the  temperature  of  18°  and  pres- 
sure of  748.4  mm.?  Ans.  3031  c.c. 

61.  10.08  grams  of  hydrogen  (mol.  wt.  =  2.016)  at  0°  and 
760  mm.  pressure  are  to  be  forced  into  a  vessel  of  11.2  liters 
capacity.     Under  what  pressure  will  the  hydrogen  be  at  this 
same  temperature?  Ans.  10  atmospheres. 

62.  At  what  temperature  will  8  grams  of  oxygen  (mol.  wt.  = 
32)  under  a  pressure  of  760  mm.  occupy  a  volume  of  11.2  liters  at 
this  same  pressure?  Ans.  273°. 

63.  A  vessel  holding  30.8  grams  of  carbon  dioxide  (mol. 
wt.  =  44)  at  10°  and  740  mm.  pressure  is  to  be  brought  to  a 
temperature  of  50°  and  a  pressure  of  750  mm.     What  weight 
and  volume  of  carbon  dioxide  will  be  lost? 

Ans.  2105  c.c.  gas  at  latter  conditions;  3.449  grams. 


AVOGADEO'S   HYPOTHESIS  45 

64.  What  is  the  weight  of  6.594  liters  of  oxygen  (mol.  wt.  = 
32)  at  40°  and  740  mm.  pressure?  Ans.  8  grams. 

65.  What  is  the  absolute  density  of  hydrogen  sulphide  (mol. 
wt.  =  34.09).  Ans.  1.52. 

66.  What  is  the  relative  density  of  hydrogen  sulphide  (mol. 
wt.  =  34.09)  referred  to  air  and  also  to  oxygen? 

Ans.    1.177  (air  =  1). 
1.065  (O  =  1). 

67.  What  is  the  relative  density  of  hydrogen  chloride  (mol. 
wt.  =  36.47)  referred  to  hydrogen,  air,  oxygen  and  chlorine? 

Ans.    18.05  (H  =  1). 

1.259  (air  =  1). 

1.139  (O  =  1). 

0.514  (Cl  =  1). 

68.  8  grams  of  oxygen  were  mixed  with  10.08  grams  of 
hydrogen  (mol.  wt.  =  2.016).     Both  gases  were  measured  at  the 
standard  conditions.     What  was  the  relative  density  of  this 
mixture  (0  =  1)?  Ans.    0.1076. 

69.  A  gas  with  the  relative  density  1.5  (0  =  1)  wag  reduced 
from  standard  conditions  to  20°  and  740  mm.  pressure  so  that 
it  might  have  a  volume  measuring  6172.3  c.c.     What  is  the 
weight  of  this  volume  of  the  gas?  Ans.  12  grams. 

70.  A  volume  of  gas,  with  the  relative  density  0.8757  (0  =  1), 
was  found  to  measure  1560  c.c.  when  transferred  to  a  vessel 
over  water  at  18°  and  742.4  mm.     What  is  the  weight  of  the 
dry  gas  here  concerned?  Ans.    1.75  grams. 


CHAPTER  VII. 
THE   LAW   OF   DEFINITE   PROPORTIONS. 

WHEN  two  or  more  elements  unite  in  chemical  combi- 
nation, a  definite  ratio  is  found  to  exist  between  the  quan- 
tities of  each  element  present  and  also  between  each  of 
these  several  quantities  and  that  of  the  molecular  weight 
of  the  compound  itself.  In  other  words,  upon  the  basis 
of  100  as  the  molecular  weight,  the  amount  by  weight, 
here  the  percentage,  of  each  element  in  the  compound  is 
constant  and  must  bear  a  constant  and  unalterable  ratio 
to  the  percentage  by  weight  of  any  other  element  present. 

The  molecular  weight  of  a  compound  represents  the 
sum  of  the  atomic  weights  that  go  to  form  it.  The  atomic 
or  unit  weight,  i.e.,  the  symbol  weight,  is  regarded  as  the 
smallest  weight  in  which  an  element  can  be  present  in  a 
molecular  weight  of  any  of  its  volatile  compounds.  A 
formula,  therefore,  is  merely  the  expression  in  symbols 
which,  under  the  existing  conditions,  denotes  the  number 
of  atomic  weights  of  the  several  elements  in  one  molecu- 
lar weight;  this  is  better  called  the  molecular  formula  (cf. 
Chap.  VIII).  The  multiples  of  these  atomic  weights 
which  occur  in  the  molecular  weight  of  a  compound  are 
always  expressed  by  integers  which  are  usually  small. 

The  Percentage  Composition  of  a  Compound.  —  The 
percentage  composition  of  a  compound  may  be  deter- 
mined by  analytical  or  synthetical  means.  Thus,  for 
example,  10  grams  of  magnesium  unite  with  6.58  grams 
of  oxygen  (when  heated  in  an  atmosphere  of  oxygen), 
and  the  resulting  magnesium  oxide  weighs  16.58  grams. 

46 


THE   LAW   OF   DEFINITE   PROPORTIONS  47 

Of  the  16.58  grams  magnesium  oxide,  10  grams  are  of 
course  magnesium,  or  the  fractional  quantity  10/16.58  rep- 
resents the  weight  of  magnesium  in  magnesium  oxide; 
the  fraction  6.58/16.58  will  represent,  then,  the  propor- 
tional amount  of  oxygen  present  in  the  compound.  These 
same  definite  ratios  hold  for  all  quantities  of  magnesium 
oxide;  hence,  if  100  grams  are  taken  as  a  basis,  10/16.58 
of  100,  or  60.3,  will  give  the  amount  of  magnesium  in 
the  100  grams,  i.e.,  the  percentage  of  this  element  in  the 
compound.  Similarly,  6.58/16.58  of  100,  or  39.7,  is  the 
percentage  of  oxygen  present.  By  reference  to  simple 
proportion  upon  the  basis  of  100  as  the  total  weight, 
these  same  percentage  amounts  are  readily  determined: 

MgO  :  Mg    =  100  :  x 

16.58  :  10     =  100  :  x,  or  60.3  per  cent 

MgO  :  O       =  100  :  x 

16.58  :  6.58  =  100  :  x,  or  39.7  per  cent. 

Now  the  formula  of  magnesium  oxide  is  MgO,  in  which 
one  atomic  weight  of  magnesium  (24.32)  is  combined  with 
one  atomic  weight  of  oxygen  (16).  From  the  Law  of 
Definite  Proportions  we  know  the  ratios  Mg/MgO  and 
O/MgO  are  constant  and,  when  expressed  in  numerical 
values  of  the  several  atomic  weights  concerned,  become 
respectively: 

Mg                    24.32  24.32 

or     — — — or 


and       MgO  24.32  +  16  40.32 

0        or      _.     16  or        16 


MgO  24.32  +  16  40.32  " 

The  constant  ratio  24.32/40.32  stands  for  the  amount  of 
magnesium  in  magnesium  oxide  in  all  amounts  of  the 
latter,  and  upon  a  percentage  basis  will  be  represented  by 
this  same  fractional  part  of  100,  i.e.,  24.32/40.32  of  100, 
or  60.3  per  cent.  Similarly,  for  the  amount  of  oxygen  in 


48  CHEMICAL   CALCULATIONS 

magnesium  oxide  when  expressed  on  a  percentage  basis, 
16/40.32  of  100,  or  39.7  per  cent,  will  be  obtained.  These 
values  are  identical  (cf.  preceding  paragraph)  with  those 
obtained  in  the  synthesis  of  this  compound  where  a  defi- 
nite weight  of  magnesium  was  converted  into  the  oxide. 
In  accordance  with  the  Law  of  Definite  .Proportions,  the 
same  relations  must  hold  for  all  determinations  in.  this 
compound. 

The  general  expression  for  problems  of  this  type  when 
the  relative  molecular  weights  are  referred  through  simple 
proportion  to  a  basis  of  100  is  as  follows: 

MgO  :  Mg      =  100  :  x 

40.32  :  24.32  =  100  :  (60.3) 
and 

MgO  :  0        =  100  :  x 

40.32  :  16       =  100  :  (39.7). 

In  the  simple  compound  water,  with  the  formula  H2O, 
the  quantity  of  hydrogen  in  one  molecular  weight  is  rep- 
resented by  two  atomic  weights  of  this  element.  This 
aggregate  of  atomic  weights,  which  represents  the  quan- 
tity in  which  any  element,  or  group  of  elements,  is  present 
in  a  molecular  weight  of  some  compound,  may  be  regarded 
as  the  formula-quantity  for  the  particular  element,  or  ele- 
ments, concerned  in  the  total  molecular  weight.  Thus 
2  H  stands  for  the  formula-quantity  of  hydrogen  (2)  in 
water  (18),  and  the  ratio  2  H/H2O  or  2/ 18  or  1/9  represents 
this  constant  proportion  between  the  formula-quantity  of 
hydrogen  and  the  molecular  weight  of  water  when  con- 
cerned, of  course,  in  this  compound.*  As  there  is  but  one 

*  The  molecular  weight  of  a  substance  is,  after  all,  a  formula- 
quantity  for  that  particular  group  of  elements  when  associated  in  this 
compound.  The  ratio  between  the  two  is  always  expressed  by  unity. 
Thus  18  is  the  formula-quantity  of  water  in  its  molecular  weight  as 
steam,  whereas  in  the  form  of  ice  the  formula-quantity,  just  as  the 
molecular  weight,  must  be  increased  to  3  (?)  times  this  number  if  the 
molecule  is  to  be  regarded  as  (H2O)3. 


THE   LAW   OF   DEFINITE   PROPORTIONS 


49 


atomic  weight  of  oxygen  in  the  molecular  weight  of  water, 
the  second  ratio  is  simply  O/H2O  or  16/18  or  8/9. 
Hence  in  water  1  part  by  weight  of  hydrogen  (1/9)  is 
combined  with  8  parts  by  weight  of  oxygen  (8/9).  Ex- 
pressed in  percentage  composition  we  have : 


H20 
18.02 


:  2  H     =  100  :  x 

:  2.02    =  100  :  11.2  per  cent  hydrogen 


and 


H2O     :  O        =  100  :  88.8  per  cent  oxygen, 

the  same  ratio,   1  :  8,  holding  true,  of  course,  for  the 
percentages. 

In  the  compound  sodium  sulphate,  Na2SO4,  we  may 
express  the  constant  ratios:  2Na/Na2S04,  S/Na2SO4,  and 
4O/Na2SO4,  by  percentages  of  the  molecular  weight, 
142.07,  or  (46.0  +  32.07  +  64),  as  follows: 


Na2S04 

2  Na  =  100 

X 

142.07 

46.0    =  100 

32.38  per  cent 

Na2S04 

S         =  100 

X 

142.07 

32.07  =  100 

22.57  per  cent 

Na2SO4 

4O      =100 

X 

142.07 

64       =  100 

45.05  per  cent 

The  sum  of  these  =  100  per  cent. 

The  Relation  of  Formula-Quantity  to  Molecular  Weight. 

-  Whatever  the  weight  of  a  substance  under  considera- 
tion may  be,  the  ratio  between  any  one  formula-quantity 
and  the  total  molecular  weight  of  the  compound  is  always 
equal  to  the  ratio  between  the  actual  weights  which  here 
correspond  to  these  respective  quantities. 

Example  19.  —  Calculate  the  weight  of  chlorine  present  in 
50  grams  of  sodium  chloride. 


50  CHEMICAL   CALCULATIONS 

The  formula  of  sodium  chloride,  NaCl,  indicates  that 
1  atomic  weight  of  sodium  (23.00)  is  combined  with 
1  atomic  weight  of  chlorine  (35.46)  to  give  a  molecular 
weight  of  58.46.  The  ratio  Cl/NaCl  or  35.46/58.46 
expresses  the  relation  between  the  formula-quantity  of 
chlorine  and  the  molecular  weight  of  the  salt,  and  all 
weights  of  these  substances  when  jointly  involved  in  the 
compound  salt  must  be  in  accord  with  this  ratio.  When 
this  ratio  is  converted  to  a  percentage  basis  we  obtain  the 
value  60.6  as  the  percentage  of  chlorine  in  all  amounts  of 
salt:  58.46  :  35.46  =  100  :  60.6.  If,  then,  60.6  per  cent  of 
any  weight  of  salt  is  chlorine,  50  grams  X  60.6  per  cent  or 
30.3  grams  is  the  amount  of  chlorine  present  in  the 
known  weight  of  salt. 

Without  resort  to  the  percentage  composition  of  a  com- 
pound the  constant  ratio  between  the  formula-quantity 
of  any  element  present  and  the  total  molecular  weight 
may  be  used  directly  for  the  determination  of  the  amount 
of  this  element  in  any  weight  of  compound.  In  the 
example  just  given,  35.46/58.46  represents  the  propor- 
tional amount  of  chlorine  in  salt.  Without  reference, 
then,  to  100  as  a  basis,  this  same  proportional  part  of  the 
weight  of  salt  taken  must  give  the  required  value  for 
chlorine;  thus  50  grams  X  35.46/58.46  =  30.3  grams  or 
weight  of  chlorine  present;  or,  as  is  more  customary,  we 
may  express  the  same  by  simple  proportion:  Since  58.46 
grams  of  salt  contain  35.46  grams  of  chlorine,  50  grams  of 
salt  can  contain  only  that  proportional  part  of  its  own 
weight  which  35.46  is  of  58.46;  the  ratio  58.46  :  35.46  at 
once  becomes  equal  to  the  ratio  50  :  x;  thus 

58.46  :  35.46  =  50  :  x,  or  x  =  30.3  grams. 

These  proportions  are  most  easily  formed  when  we 
bear  in  mind  that  only  one  set  of  related  terms,  e.g.,  the 


THE   LAW   OF   DEFINITE   PROPORTIONS  51 

formula-quantities,  constitute  one  member  of  the  propor- 
tion, whereas  the  other  set  of  related  terms,  the  actual 
weights  (here  with  one  unknown  factor),  constitute  the 
other  member. 

Example  20.  —  Calculate  the  weight  of  sodium  chloride 
that  can  be  prepared  from  30.3  grams  of  chlorine. 

The  amount  of  salt  that  can  be  prepared  from  this 
definite  weight  of  chlorine  (30.3  grams)  will  be  represented 
by  the  same  ratio,  Cl/NaCl,  as  under  the  preceding 
example  (Ex.  19);  but,  for  the  purposes  of  multiplication 
into  known  values,  it  is  necessary  to  use  the  inverted 
form  NaCl/Cl  or  58.46/35.46,  with  the  large  value  in  the 
numerator  so  that  a  proportional  increase  in  the  values 
affected  by  this  ratio  may  be  obtained.  The  same  end 
may  be  reached  by  the  use  of  simple  proportion,  where 
the  unknown  term  increases  over  the  known  in  direct 
ratio  to  the  corresponding  formula-quantities : 

35.46  :  58.46  =  30.3  :  x,  or  x  >=  50  grams. 

The  Relation  of  Formula-Quantities  to  Each  Other.  — 

The  ratio  between  any  two  formula-quantities  occurring 
in  one  molecular  weight  may  be  used,  of  course,  inde- 
pendently of  the  molecular  weight,  and  when  the  ratio 
between  certain  groups  of  elements  in  a  formula  is  to  be 
determined  the  same  principle  will  be  found  to  apply 
as  between  single  elements.  The  symbols  representing 
these  groups  serve  as  usual  for  the  calculation  of  the 
correct  formula-quantities.  In  analyses  of  minerals  the 
determination  of  these  formula-quantities  for  various 
groups  of  oxides,  etc.,  aids  materially  in  the  general  classi- 
fication; thus  in  the  mineral  feldspar,  A1K  (Si3O8),  or 
Al2K2(Si3O8)2,  when  intended  for  resolution  into  the  con- 
stituent groups,  becomes  A1203 .  K2O  .  6  Si02,  from  which 


52  CHEMICAL   CALCULATIONS 

the  percentage  of  any  one  of  the  oxides  present  may  be 
calculated  from  the  corresponding  formula-quantity. 

Example  21. — What  weight  of  aluminium  is  present  in  a 
sample  of  pure  cryolite,  A12F6 .  6  NaF,  which  analyzed  for  10 
grams  of  sodium? 

The  ratio  between  the  formula-quantity  of  each  of  the 
two  substances  concerned  is  2  Al/6  Na,  or  2  (27.1)/6  (23), 
or  54.2/138.  As  this  is  a  constant  we  need  only  refer  the 
ratio  directly  to  the  equivalent  ratio  when  based  upon 
sodium  as  10  grams,  — 54.2/138  =  x/ 10,  —  or,  as  has  been 
the  practice,  138  :  54.2  =  10  :  x,  from  which  we  ob- 
tain the  value  3.93  grams  for  x,  the  weight  of  aluminium 
present. 

The  Relation  of  Formula-Quantities  to  Percentage  Com- 
position. —  When  data  are  given  in  percentages  and  it 
is  desired  to  ascertain  the  percentage  amount  of  some  other 
element  or  group  of  elements  present  in  the  same  com- 
pound, the  calculations  are  made  directly  upon  these  per- 
centages. They  occupy,  of  course,  the  identical  relation 
to  each  other  as  do  the  actual  weights  themselves,  but 
with  the  condition  that  a  constant  value  of  100  serves  as  a 
basis  for  the  entire  molecular  weight. 

Example  22.  —  What  percentage  of  potassium  oxide,  K2O,  is 
present  in  a  sample  of  feldspar,  AlK(Si308),  which  analyzed 
for  12.2  per  cent  potassium? 

The  ratio  between  the  potassium  and  its  corresponding 
amount  of  potassium  oxide  possible  of  existence  in  this 
same  molecule  is  expressed  by  2  K/K2O.  With  this  defi- 
nite quantity  of  oxygen  in  the  molecule  now  to  be  asso- 
ciated with  the  potassium,  there  naturally  will  be  an 
increase  in  the  percentage  amount  of  the  group  K2O  over 
that  of  the  potassium  alone.  The  ratio  between  these 
percentage  amounts  must  be  in  accordance  with  the  corre- 


THE   LAW   OF   DEFINITE   PROPORTIONS  53 

spending  formula-quantities,  2  K  and  K2O,  upon  which 
their  weight  and  consequent  percentage  is  dependent; 
therefore  the  ratio  2  K/K2O  must  be  equal  to  the  ratio 
12.2% lx%.  From  the  equation  2  K/K2O  or  78.2/94.2  = 
12. 2 /x  we  obtain,  as  the  value  of  x,  14.7  per  cent,  i.e., 
14.7  per  cent  of  a  compound  of  the  composition  noted, 
and  analyzing  for  12.2  per  cent  potassium,  may  be  con- 
sidered as  potassium  oxide. 

Percentage  Composition  as  a  Basis  for  Percentage 
Purity.  —  Upon  this  method  of  comparison  it  is  a  very 
simple  task  to  calculate  the  percentage  purity  or  per- 
centage amount  of  some  compound  in  one  of  its  samples 
submitted  to  an  analysis;  and  this  too  upon  the  data 
secured  in  the  determination  of  only  one  of  its  constituent 
elements. 

Example  23.  —  A  sample  of  salt  analyzed  for  50  per  cent 
chlorine.  Assuming  all  of  the  chlorine  to  have  been  present  as 
sodium  chloride,  what  was  the  percentage  of  salt  in  the  sample? 

The  ratio  Cl/NaCl  must  be  equal  to  the  ratio  between  the 
corresponding  percentage  amounts  of  these  quantities, — 
50% /z%.  Cl/NaCl  =  35.46  :  58.46  =  50/z  or,  by  simple 
proportion,  35.46  :  58.46  =  50  :  x.  From  this,  x  is  found 
to  be  82.4  in  place  of  the  theoretical  100.  This  is,  there- 
fore, the  percentage  purity  of  the  sample.  When  analyti- 
cal data  are  not  given  in  percentages,  the  weight  of  both 
sample  and  final  product  must  be  known  in  order  to  deter- 
mine the  percentage  amount  of  any  constituent  present. 

The  percentage  purity  of  a  compound  may  be  deter- 
mined also  by  calculating  the  weight  of  substance  theo- 
retically possible  from  the  results  at  hand  (cf.  Ex.  20) 
and  then  determining  what  percentage  the  weight  of 
sample  bears  to  this  calculated  value. 

The  Interrelationship  of  two  Formula-Quantities  through 
a  Common  Quantity.  —  When  the  results  of  an  analysis 


54  CHEMICAL   CALCULATIONS 

are  given  in  terms  of  some  compound  other  than  that 
involved  in  the  investigation  it  is  necessary  to  calculate 
these  results  through  the  element  common  to  both  com- 
pounds. 

Example  24.  —  A  sample  of  feldspar,  KAlSi308,  weighing 
0.2507  gram  gave  upon  analysis  0.0658  gram  of  potassium 
sulphate,  K2S04.  What  was  the  percentage  of  potassium  oxide 
present  in  the  sample? 

We  have  here  two  distinct  ratios:  K2S04/2  K,  which 
answers  for  the  definite  quantity  of  potassium  present  in 
this  weight  of  potassium  sulphate,  and  also  2  K/K2O 
for  the  quantity  of  oxide  possible  from  a  definite  weight 
of  potassium  (cf.  Ex.  22).  Since  the  ratios  involve  the 
same  definite  quantities  of  potassium  throughout,  they 
may  be  considered  as  equivalent  in  respect  to  this  ele- 
ment, and  when  brought  together  may  be  cleared  of  this 
common  term,  2  K,  as  indicated  when  K2SO4/2  K  is 
involved  with  2  K/K2O.  By  multiplication  we  should 
have  K2S04/K2O,  or  that  ratio  with  which  the  corre- 
sponding ratio  between  the  actual  weights  of  these 
substances,  involving  the  same  weight  of  potassium, 
must  accord.  From  the  known  weight  of  potassium  sul- 
phate, the  actual  weight  of  potassium  oxide  is  easily 
calculated:  K2SO4/K2O  or  174.27/94.27  =  0.0658/z,  i.e., 
174.27  :  94.27  =  0.0658  :  x,  or  x,  the  weight  of  potassium 
oxide  in  the  sample,  is  equal  to  0.0356  gram.  From  this 
weight  of  oxide  and  the  original  weight  of  the  sample  the 
percentage  amount  of  potassium  oxide  in  the  compound 
may  be  calculated:  0.2507  :  0.0356  =  100  :  x,  orx=  14.2 
per  cent.  Without  the  simplification  noted  it  would 
have  been  necessary  first  to  determine  the  actual  weight 
of  potassium  present  in  the  potassium  sulphate,  and 
second,  from  the  weight  of  potassium,  the  consequent 
weight  of  oxide. 


THE   LAW    OF   DEFINITE   PROPORTIONS  55 

This  cancellation  or  elimination  of  a  common  term 
between  two  ratios  may  be  illustrated  by  a  second  example 
somewhat  more  complicated. 

Example  25. — What  percentage  of  potassium  chloride  can 
be  regarded  as  present  in  a  sample  of  potassium  chlorplatin- 
ate,  K2PtCl6,  which  analyzed  for  43.6  per  cent  chlorine? 

The  ratio  of  the  formula-quantity  of  chlorine  to  that  of 
potassium  is  6  Cl/2  K.  Now  the  ratio  between  potassium 
and  potassium  chloride  in  one  molecular  weight  of  potas- 
sium chloride  is  K/KC1.  '  By  examination  of  the  formula 
of  potassium  chlorplatinate  it  is  seen  that  two  molecules 
of  potassium  chloride  are  involved  in  each  molecule  of 
chlorplatinate.  This  latter  ratio  then  becomes  2  K/2  KC1 
when  considered  in  the  molecule  of  chlorplatinate.  By 
this  manner  the  same  formula-quantity  of  potassium 
is  considered  in  both  ratios  (6  Cl/2  K  and  2  K/2  KC1); 
consequently  this  equivalent  quantity  may  be  eliminated 
from  each  (cf.  Ex.  24),  and  the  final  ratio  6  Cl/2  KC1  ob- 
tained for  the  corresponding  weights  of  chlorine  and  potas- 
sium chloride  possible  in  a  molecule  of  chlorplatinate. 
Though  a  fractional  quantity  of  the  total  chlorine  is 
brought  into  consideration  as  potassium  chloride,  this  in 
no  way  disturbs  the  definite  relation  between  the  total 
chlorine  and  the  total  potassium  even  when  considered 
as  chloride.  The  ratio  may  be  simplified,  if  desired, 
to  3  C1/KC1,  and  is  equal  to  the  ratio  between  the  cor- 
responding percentage  amounts  of  these  substances, 
43.6%  /x%.  From  the  equation 

.  3  C1/KC1  =  106.38  :  74.56  =  43.6/z 

we  obtain  30.56  as  the  percentage  amount  of  potassium 
chloride  possible  in  the  sample. 


56  CHEMICAL   CALCULATIONS 


PROBLEMS. 

71.  Determine  the  percentage  composition  of  calcium  car- 
bonate, CaC03.         Ans.    Ca,  40.05% ;  C,  11.99% ;  0,  47.96%. 

72.  Determine  the  percentage  composition  of  sodium  nitrite, 
NaN02.  Ans.    Na,  33.33%;  N,  20.30%;  0,  46.37%. 

73.  Determine   the   percentage    composition   of   potassium 
chlorate,  KC1O3,       Ans.   K,  31.90%;  Cl,  28.93%;  O,  39.17%. 

74.  Determine  the  percentage  composition   of   crystallized 
sodium  sulphate,  Na2S04 . 10  H2O. 

Ans.   Na,  14.28%;  S,  9.95%;  0,  69.51%;  H,  6.26%. 

75.  Determine  the  percentage  composition  of  sulphuric  acid, 
H2SO4.  Ans.    H,  2.05%;  S,  32.70%;  O,  65.25%. 

76.  Determine  the  percentage  of  sulphur  trioxide,  SO3,  in 
sulphuric  acid.  Ans.    81.64  per  cent. 

77»    Determine  the  percentage  of  water  in  crystallized  sodium 
carbonate,  Na2C03 .  10  H2O.  Ans.    62.95  per  cent. 

78.  Determine  the  percentage  composition  of  ordinary  alum, 
K1A12(SO4)4.24H20. 

Ans.  K,  8.24%;  Al,  5.71%;  S,  13.52%;  0,  67.43%;  H,  5.10%. 

79.  What  weight  of  sodium  is  present  in  50  grams  of  sodium 
nitrate,  NaNO,?  Ans.    13.53  grams. 

80.  What  weight  of  sodium  is  present  in  100  grams  of  sodium 
hydrogen  carbonate,  NaHC03?  Ans.    27.38  grams. 

81.  What  weight  of  oxygen  is  present  in  200  grams  of  potas- 
sium chlorate,  KC1O3?  Ans.    78.34  grams. 

82.  What  weight  of  oxygen  is  present  in  200  grams  of  mer- 
curic oxide,  HgO?  Ans.    14.815  grams. 

83.  What  weight  of   mercuric  oxide,  HgO,  will  contain  30 
grams  of  oxygen?  Ans.    405  grams. 

84.  What  weight  of  potassium  chlorate  will  contain  30  grams 
of  oxygen?  Ans.    76.6  grams. 

85.  What  weight  of  sulphuric  acid  can  be  prepared  from 
100  grams  of  sulphur?  Ans.    305.9  grams. 

86.  Calculate  the  weight  of  potassium  in  a  sample  of  pure 
sylvite,  KC1,  which  analyzed  for  2.230  grams  of  chlorine. 

Ans.    2.459  grams. 


THE   LAW   OF   DEFINITE   PROPORTIONS  57 

87.  What  weight  of  copper  is  present  in  a  sample  of  pure 
copper  sulphate,  CuS04 .  5  H20,  which  analyzed  for  30.2  grams 
of  sulphur  trioxide,  SO3?  Ans.    23.97  grams. 

88.  Calculate  the  percentage  purity  of  a  sample  of  horn- 
silver,  AgCl,  which  analyzed  for  74.2  per  cent  silver. 

Ans.    98.61  per  cent. 

89.  Calculate  the  percentage  purity  of  a  sample  of  marble, 
CaC03,  which  analyzed  for  39.6  per  cent  calcium. 

Ans.    98.88  per  cent. 

90.  Calculate  the  percentage  of  potassium  chloride  in  a  sample 
of  carnallite,  KC1 .  MgCl2 .  6  H2O,  which  analyzed  for  37.72  per 
cent  chlorine.  Ans.    26.44  per  cent. 

Note.  —  Theory  calls  for  26.83  per  cent  KC1.  Therefore  the  devia- 
tion; 0.39,  between  the  percentage  values  means  an  error  of  1.5  per  cent 
from  the  theory  (26.83).  Halogen  determinations  are  usually  quite 
accurate.  If  the  analysis  is  at  all  good,  the  salt  falls  a  little  short  of 
purity. 

91.  Calculate  the  percentage  of  calcium  oxide,  CaO,  present 
in  a  sample  of  marble,  CaC03,  which  analyzed  for  43.8  per  cent 
carbon  dioxide.  Ans.    55.84  per  cent. 

Sample  practically  pure;  theory,  56.04  per  cent. 

92.  A  sample  of  sodium  chromate,  Na2CrO4J  weighing  1.6780 
grams,  gave  upon  analysis  1.4620  grams  of  sodium  sulphate, 
Na2SO4.     What  was  the  percentage  of  sodium  oxide,  Na2O,  in 
the  sample?  Ans.    38.01  per  cent. 

Theory,  38.25  per  cent. 

93.  Calculate  the  percentage  purity  of  a  quantity  of  potas- 
sium ferrocyanide,  K4Fe(CN)6,  0.5793  gram  of  which  gave  upon 
analysis  0.4650  gram  of  potassium  sulphate,  K2S04. 

Ans.    84.8  per  cent. 

Note.  —  From  this  weight  of  K2S04  determine  the  actual  weight, 
and  finally  the  percentage,  of  potassium  in  the  sample  and  then  apply 
Ex.  23. 

94.  What  is  the  percentage  of  potassium  sulphate,  K2SO4,  in 
a  sample  of  common  alum,  K2S04 .  A12(S04)3 .  24  H20,  which 
analyzed  for  33.51  per  cent  sulphur  trioxide,  SO3? 

Ans.    18.23  per  cent  K3SO4. 
Sample  practically  pure;  theory,  18.36  per  cent  K,SO4. 


58  CHEMICAL   CALCULATIONS 

95.  What  is  the  percentage  of  copper  carbonate,  CuC03,  in 
a  sample  of  malachite,  Cu(OH)2 .  CuCO3,  which  analyzed  for  57.1 
per  cent  copper?  Ans.    55. 49  per  cent  CuCO3. 

The  salt  is  probably  pure;  theory,  55.87  per  cent  CuCO8. 

96.  A  sample  of  carnallite,  KC1 .  MgCl2 .  6  H2O,  analyzed  for 
35.34  per  cent  chlorine.     What  is  the  percentage  of  magnesium 
chloride  present?  Ans.    31.64  per  cent  MgCl2. 

Note.  — Theory:  34.27  per  cent  MgCl2.     Therefore  92.31  per  cent 
pure. 

97.  A  sample  of   crystallized  ferrous-ammonium  sulphate, 
FeSO4 .  (NH4)2S04 .  6H2O,  analyzed  for  8.66  per  cent  ammonia, 
NH3.     What  is  the  percentage  of   ferrous   sulphate,  FeSO4, 
present?  Ans.    38.60  per  cent  FeSO4. 

Theory,  38.74  per  cent  FeSO4. 

98.  A  sample  of  gahnite,  Zn(A102)a  or  ZnO  .  A12O3,  weighing 
0.1909  gram,  gave  on  analysis  0.1664  gram  of  zinc  sulphate, 
ZnS04.    What  was  the  percentage  of  alumina,  A1203,  in  the 
sample?  Ans.    55.18  per  cent. 

If  analysis  is  correct,  the  sample  is  slightly  impure. 

Theory,  55.67  per  cent. 


CHAPTER  VIII. 
THE    DERIVATION    OF    CHEMICAL   FORMULAE. 

Molecular  Formulae  from  Molecular  Weight  and  Per- 
centage Composition.  —  The  formula  of  a  compound, 
designated  by  symbols  which  stand  individually  for  one 
atomic  weight  of  an  element,  is  derived  from  data  afforded 
by  analysis  or  synthesis.  In  reverse  to  the  determination 
of  the  percentage  composition,  as  described  in  the  preced- 
ing chapter,  we  can  derive  the  true  molecular  formula 
when  given  the  correct  molecular  weight  of  a  compound 
and  the  percentage  amount  of  each  element  present. 
The  proportional  parts  of  the  molecular  weight  indicated 
by  the  several  percentages  are  of  course  the  quantities 
(formula-quantities)  of  the  respective  elements  in  one 
molecular  weight  of  the  compound,  and  necessarily  must 
be  multiples  of  the  corresponding  atomic  weights  by  unity 
or  a  small  integer. 

Example  26.  —  The  molecular  weight  of  water  is  18.016.  The 
amount  of  hydrogen  present  is  11.2  per  cent  and  that  of  oxygen 
88.8  per  cent.  What  is  the  molecular  formula? 


Proportional 

Unit  or 

Molecular 
weight. 

Percentage 
indicated. 

part  of  molec- 
ular weight 
corresponding 

atomic 
weight  of 
element. 

Number  of 
units. 

to  element. 

18.016 

H=  11.2% 

2.016 

1.008 

2 

18.016 

O=  88.8% 

16.00 

16.00 

1 

59 


60  CHEMICAL   CALCULATIONS 

The  values  2.016  and  16  represent  respectively  the  sum  of 
the  atomic  weights  of  hydrogen  and  oxygen  which  are 
present  in  the  molecular  weight  of  water.  They  are, 
therefore,  the  formula-quantities  of  these  elements  in 
this  compound.  With  1.008  and  16  as  the  atomic  weights 
of  hydrogen  and  oxygen  respectively,  it  is  only  a  simple 
step  to  determine  the  number  of  hydrogen  units  (2)  and 
oxygen  units  (1)  which  are  necessary  to  make  up  the 
formula-quantity  of  each  element  in  this  molecule  and 
consequently,  together,  the  molecular  weight  of  water; 
hence  the  formula  H20. 

The  correct  molecular  weight  of  a  compound  is  rarely 
ever  at  hand  for  the  determination  of  chemical  formulae. 
The  usual  procedure  lies,  then,  in  the  determination  of 
the  formulae  from  the  percentage  composition  or  other 
data,  and  in  the  final  adjustment  of  these  to  accord  with 
the  molecular  weights  which  may  have  been  determined 
only  with  approximate  exactness. 

Empirical  Formulae  from  Percentage  Composition. — 
Upon  the  percentage  basis  the  molecular  weight  is  re- 
garded as  brought  over  to  the  value  100;  this  number, 
however,  gives  in  itself  no  clue  to  the  probable  molecular 
weight.  In  Example  26  we  were  given  the  correct  molec- 
ular weight;  consequently,  the  proportional  parts  of  this 
total  weight,  as  indicated  by  the  percentage  amounts, 
agree  always  with  the  exact  quantity  of  each  element 
present  in  one  molecular  weight  of  the  compound.  Now, 
upon  the  adoption  of  100  as  the  molecular  weight,  each 
percentage  amount  becomes  the  accepted  value  for  the 
formula-quantity  of  that  particular  element  in  the  molec- 
ular weight.  In  dividing  each  of  these  values  through  by 
the  atomic  weight  of  the  corresponding  element,  we  obtain 
not  whole  numbers,  indicating  the  number  of  atomic 
weights  of  the  respective  elements  present,  but  fractions 


DERIVATION   OF   CHEMICAL   FORMULA  61 

or  factors  of  these  whole  numbers,  all  of  which  are 
related  to  the  true  numbers  in  the  same  ratio  as  the 
adopted  value,  100,  is  related  to  the  true  molecular 
weight. 

Since  a  chemical  formula  calls  for  simple  multiples  of 
the  atomic  weights  concerned,  we  need  only  raise  the 
entire  range  of  factors  by  some  term  which  will  bring 
each  and  all  into  whole  numbers.  When  the  smallest 
factor  is  made  the  divisor,  then  all  of  the  other  factors 
divided  through  by  it  must  necessarily  give  quotients 
which  are  equal  to  or  greater  than  this  smallest  factor  as 
unity.  In  bringing  the  correct  molecular  weight  over  to 
100  as  a  basis,  this  smallest  factor  was  of  course  reduced 
from  unity,  or  a  multiple  of  it,  in  the  same  proportion  as 
were  all  of  the  other  factors.  The  quotients,  then,  upon 
the  basis  of  this  small  factor  as  unity,  will  possess  values 
close  to  their  former  and  correct  numbers.  If  the  smallest 
factor  had  been  reduced  from  unity  itself,  then  the  quo- 
tients will  represent  the  correct  formula,  —  a  molecular 
formula,  —  of  the  compound.  If,  however,  the  original 
value  of  this  smallest  factor  was  a  simple  multiple  of  unity 
(e.g.,  3),  then  the  quotients  will  vary  from  their  true  values 
in  a  molecular  formula  by  just  this  same  fractional  amount 
that  unity  is  of  the  simple  multiple.  (If  3  were  the  mul- 
tiple, then  the  quotients  would  stand  at  one-third  of  their 
original  values.)  This  simplest  expression  of  a  formula  in 
symbols  is  known  as  an  empirical  formula,  and  the  sum  of 
the  atomic  weights  therein  represented,  though  a  formula 
weight,  is  not  necessarily  the  molecular  weight. 

From  the  empirical  formula  the  true  percentage  com- 
position of  a  compound  is  always  derivable.  Its  formula 
weight,  however,  may  not  be  coincident  with  the  molecular 
weight,  and,  if  so,  the  correct  molecular  formula  can  be 
derived  only  when  the  molecular  weight,  or  at  least  a  fair 


62 


CHEMICAL   CALCULATIONS 


approach  to  this,  is  known.  The  molecular  formula,  there- 
fore, is  always  a  multiple  of  the  empirical  formula  by  some 
small  integer. 

Example  27.  —  A  substance  by  analysis  was  found  to  contain 
32.32  per  cent  sodium,  22.44  per  cent  sulphur,  and  45.24  per 
cent  oxygen.  What  is  the  formula  of  the  compound? 

The  method  of  solution  will  be  made  most  apparent 
when  we  set  down  the  percentages  and  refer  these  to  the 
respective  atomic  weights  of  the  elements  present,  as 
indicated  below: 


Percentage 

Substance. 

Percentage. 

referred  to 
corresponding 

Factor. 

Factor  to  unit 
value. 

Sodium  .  . 
Sulphur.. 
Oxygen.  . 

at.  wt.  as  basis 

32.32 
22.44 
45.24 

32.32/23.00 
22.44/32.07 
45,24/16 

1.406  ]  Divide      [2.01 
0.699  [  through  \   1.00 
2.827  J  by  0.699  I  4.04 

100.00 

These  unit  values  approach  very  closely  to  the  integers 
2,  1  and  4,  and  designate  the  number  of  unit  or  atomic 
weights  of  sodium,  sulphur  and  oxygen,  respectively,  in 
their  corresponding  formula-quantities  in  a  formula 
weight  of  the  compound,  Na2SO4. 

In  Chapter  VII  we  have  noted  the  means  for  calculating 
the  theoretical  percentage  composition  of  a  compound. 
In  the  case  of  a  derived  formula  it  is  always  well  to  calcu- 
late from  it  the  percentage  amount  of  each  element  present 
and  note  whether  or  not  these  values  check  with  those 
derived  by  analysis.  The  derived  formula  Na2SO4  presents 
the  following  percentage  composition: 


DERIVATION   OF   CHEMICAL   FORMULAE  63 


Na2SO4 
142.07 

Na2S04 
142.07 

Na2SO4 
142.07 


2Na 
46 


£  =  100  :  x,  or  32.38  per  cent  sodium. 


32 


4  O 
64 


) 
07  \ 

) 

\ 
' 


*  X'  °r  22'5^  Per  cent 


:  x,  or  45.05  per 

100       per  cent. 


The  percentage  amounts  actually  found  by  analysis 
approach  very  closely  to  these  theoretical  values  based 
upon  the  formula  Na2SO4;  consequently  we  may  conclude 
that  this  formula  represents  the  constitution  of  the  com- 
pound. The  molecular  weight  of  sodium  sulphate  is 
found  to  be  close  to  142;  hence  the  formula  Na2S04 
(with  the  formula  weight  2  (23)  +  32.07  +  4  (16)  or 
142.07)  satisfies  also  the  requirements  for  a  molecular 
formula. 

Accuracy  in  analytical  data  is  dependent  upon  purity 
of  material  and  method  of  operation.  In  all  of  our 
results  we  may  expect  a  certain  degree  of  variation  from 
the  theoretical  values,  but  the  limits  of  error  in  the  deter- 
mination of  chemical  formulae  should  not  greatly  exceed 
two-tenths  of  1  per  cent.  Much  depends,  however,  upon 
the  particular  element  considered.  For  example,  in  the 
case  of  hydrogen  the  error  often  may  be  as  much  as  four- 
tenths  of  1  per  cent  above  the  theoretical,  due  to  insuffi- 
cient removal  of  moisture  from  the  sample  or  to  a  faulty 
combustion.  In  the  case  of  oxygen  the  data  are  rarely 
ever  determined  directly,  but  by  difference;  thus  in 
Example  27  the  sum  of  the  percentage  amounts  for 
sodium  and  sulphur  was  subtracted  from  the  total  100 
per  cent,  and  the  difference  considered  as  the  value  for 
oxygen,  a  value  altogether  dependent  upon  the  accuracy 


64  CHEMICAL   CALCULATIONS 

of  the  other  data.  From  these  considerations  the  chemist 
finds  it  greatly  expedient,  in  problems  of  this  nature,  to 
consider  each  analysis  separately  and  to  determine  what 
one  is  likely  to  be  most  free  of  error.  The  factor,  there- 
fore, which  corresponds  to  the  percentage  amount  of  this 
one  element  is  the  best  to  select  as  a  basis  for  the  reduction 
of  all  the  other  factors  to  unit  values.  Let  us  suppose, 
in  Example  27,  that  the  value  for  sodium  was  known  to 
be  more  accurately  determined.  The  factor  here  is  1.406, 
and,  if  we  base  all  the  other  data  upon  this  one,  i.e., 
by  dividing  each  factor  by  1 .406,  we  shall  obtain  the  values, 
1,  0.498  and  2.010,  for  sodium,  sulphur  and  oxygen  respec- 
tively. These  may  be  brought  to  unit  values  by  multi- 
plication by  a  small  integer,  —  here  by  2,  —  and  we  obtain 
finally  2,  0.996  and  4.02,  —  values  which  approach  the 
respective  integers  2,  1  and  4  somewhat  more  closely  than 
in  the  previous  calculation.  This,  of  course,  is  due  to  the 
close  agreement  between  the  theoretical  and  found  per- 
centage values  for  sodium.  It  is  now  seen  why  the  factor 
for  oxygen  is  rarely  selected  as  a  basis  for  formula  deter- 
minations. 

Example  28.  —  Determine  the  formula  for  that  substance 
which  presented,  by  analysis,  the  following  percentage  com- 
position: carbon,  39.78  per  cent,  hydrogen,  6.97  per  cent  and 
oxygen,  53.25  per  cent.  Above  230°  the  vapor  of  this  sub- 
stance gave  a  constant  relative  density,  2.09  (air  =  1),  calcu- 
lated to  standard  conditions. 

From  the  relative  density  it  is  only  a  simple  step  to 
calculate  the  molecular  weight  (cf.  Ex.  14): 
1  :  2.09  =  28.955  :  x, 

where  x,  the  molecular  weight,  is  found  to  be  60.52. 

The  following  table  of  percentages  and  corresponding 
atomic  weights  is  easily  constructed  as  under  Example  27: 


DERIVATION    OF    CHEMICAL    FORMULAE 


65 


Percentage 

Substance. 

Percentage. 

referred  to 
corresponding 

Factor. 

Unit  value. 

at.  wt.  as  basis. 

Carbon  .... 
Hydrogen  . 
Oxygen  

39.78 
6.97 
53.25 

39.78/12 
6.97/1.01 
53.25/16 

3.315  ]  Divide       f  1.00 
6.90     |  through    |   2.08 
3.33    J  by  3.315    I  1.01 

These  unit  values  approach  well  the  integers  1,  2  and  1, 
which  represent  now  the  number  of  atomic  weights  of  these 
respective  elements  in  a  formula  weight  of  the  compound, 
CH2O.  The  sum  of  the  formula-quantities  thus  repre- 
sented in  the  formula  weight  is  30.02  (i.e.,  12  +  2.02  +  16), 
a  value  which  does  not  coincide  with  the  molecular  weight 
calculated  from  the  vapor-density  determination  above 
(60.52).  The  molecular  formula  as  previously  stated  is 
always  a  multiple  of  this  simplest  or  empirical  formula, 
and  by  just  that  integer  which  brings  the  empirical  formula 
weight  up  to  the  molecular  weight  or  an  approximation 
to  the  same.  The  integer  in  the  present  case  is  60.52/30.02 
or  2.02.  This  is  well  within  the  limits  of  error,  which  in 
some  cases  may  exceed  a  variation  of  0.2  from  an  integral 
value.  The  empirical  formula  CH2O  must  be  multiplied 
now  by  the  integer  2,  when  we  shall  obtain  C2H4O2,  the 
correct  molecular  formula,  or  that  formula  in  which  the 
sum  of  the  atomic  weights  involved  gives  the  molecular 
weight.  The  calculation  of  the  theoretical  percentage 
composition  for  this  substance,  acetic  acid,  gives  the 
following:  carbon,  39.97  per  cent,  hydrogen,  6.73  per  cent 
and  oxygen,  53.50  per  cent.  The  analytical  data,  there- 
fore, are  closely  in  accord  with  the  theoretical. 

Formulae  from  Percentage  Composition  involving  Radi- 
cals. —  As  a  more  complex  example  it  will  be  well  to 


66  CHEMICAL   CALCULATIONS 

examine  the  analytical  data  upon  some  mineral  on  which 
the  results  are  given  in  percentage  amounts  of  the  several 
groups  present.  Feldspar  belongs  to  this  class  of  sub- 
stances, and~  in  its  analyses  certain  values  will  be  found 
for  silicon  dioxide  (SiO2),  aluminium  oxide  (A1203)  and 
potassium  oxide  (K2O);  but  associated  with  these  there 
may  occur  other  oxides  which  are  isormorphous  with  one 
or  more  of  the  above.  Consequently  the  factor  for  these 
isomorphous  substances,  found,  of  course,  as  heretofore 
described,  by  the  relation  of  percentage  composition  to 
atomic  weight  or  groups  of  atomic  weights  representing  the 
substance,  must  be  reckoned  together  as  one  factor  and 
hence  as  one  group,  in  order  to  determine  the  extent  to 
which  this  particular  group  of  isomorphous  substances 
may  be  present  in  the  complete  formula.  If,  perchance, 
a  portion  of  the  data  were  not  given  in  terms  of  the 
proper  isomorphous  compound,  but  in  some  part  of  it,  for 
instance,  the  percentage  of  potassium  instead  of  potassium 
oxide,  we  should  require  only  a  single  calculation  to  con- 
vert the  data  into  the  correct  form  (cf.  Ex.  22) : 

2  K  :  K20  =  —  per  cent  given  :    —  per  cent  sought. 

Example  29.  —  A  sample  of  the  mineral  orthoclase  (a  feld- 
spar) gave  by  analysis  the  following  percentage  composition: 
Si02,  65.69  per  cent,  A1203,  17.97  per  cent,  (CaO,  1.34  per 
cent,  K2O,  13.99  per  cent,  Na20j  1.01  per  cent,  —  isomor- 
phous). What  is  its  formula? 

By  drawing  up  a  table  of  these  percentages  and  referring 
each  to  the  formula  weight  of  the  respective  groups,  we 
may  expect  to  obtain  factors  which  in  this  case  do  not 
refer  to  the  simplest  relation  between  the  elements,  but 
rather  between  the  groups  of  elements. 

The  factors  for  K2O,  Na2O  and  CaO,  since  these  oxides 
are  isomorphous  and  mutually  may  replace  each  other  in 


DERIVATION   OF   CHEMICAL   FORMULAE 


67 


the  molecule,  must  be  added  together  and  considered  as 
some  generic  formula  such  as  R2O,  when  in  the  final 
adjustment  of  the  number  of  each  group  present  in  the 
formula  they  will  figure  as  an  individual  group  of  uniform 
composition. 


Percentage  re- 

Sub- 
stance. 

Per- 
centage. 

referred  to 
formula 
wt.  of  group 

Factor. 

Factor  of  isomor- 
phous  groups. 

Unit 
value. 

as  basis. 

SiO2.  . 

65.69 

65.69/60.3 

1.0894 

1.0894 

6.258 

A1A. 

17.79 

17.79/102.2 

0.1741 

0.1741 

Divide 

1.000 

CaO.. 

1.34 

1.34/56.09 

0.02391 

through  • 

K2O.. 

13.99 

13.99/94.2 

0.1485J 

0.1887 

by 

1.084 

Na-^O. 

1.01 

1.01/62.2 

0.0163J 

0.1741 

As  has  been  already  noted  in  selecting  the  factor  of 
those  equally  small,  for  division  into  the  entire  range  of 
factors,  it  is  always  better  to  select  that  one  which  is 
likely  to  have  been  based  upon  more  accurate  analytical 
results,  —  in  this  case  the  aluminium  oxide  and  not  the 
composite  oxide  (into  the  determination  of  which  three 
analyses  must  have  entered).  The  resulting  integers  are 
here  found  to  be  6,  1  and  1,  and  consequently  the  formula 
for  the  feldspar  is  (K2O  .  CaO  .  Na20)  (A12O3)  6  (Si02),  in 
which  the  oxides  of  the  metals  K,  Na  and  Ca  make  up  one 
single  group  of  the  general  formula  of  a  feldspar,  — 
(K2O)  .  (A1203)  .  6  (Si02),  or  KAlSi308.  The  extent  of 
these  isomorphous  replacements  within  any  group  may 
vary  considerably.  This  empirical  formula  is  accepted 
as  the  true  formula  in  the  absence  of  any  conflicting 
statements  in  regard  to  the  true  molecular  weight. 

Formulae  from  the  Relation  of  Formula-Quantities  to 
Each  Other.  —  Without  recourse  to  percentage  composi- 


68  CHEMICAL   CALCULATIONS 

tion,  the  actual  weight  of  any  element  found  by  analysis 
to  be  present  in  a  known  weight  of  compound  may  be 
referred  directly  to  the  formula-quantity  of  this  element 
in  the  total  molecular  weight  of  that  compound.  Since 
these  analytical  data  are  dependent  upon  the  number  of 
atomic  weights  of  each  element  present  in  one  moleculai 
weight ,  the  ratio  between  the  weights  of  each  element  in  a 
known  weight  of  compound  containing,  for  example,  two 
constituents,  must  be  equal  to  the  ratio  between  a  certain 
unknown  number  of  atomic  weights  of  the  one  element 
and  that  of  the  other,  i.e.,  to  the  ratio  between  the  cor- 
responding formula-quantities. 

Example  30.  —  Determine  the  formula  of  an  iron  oxide  pro- 
duced in  the  oxidation  of  22.4  grams  of  iron  to  a  final  weight  of 
32  grams. 

Here  32  —  22.4  =  9.6  grams  of  oxygen  taken  up.  The 
ratio  between  the  weights  of  these  two  constituents  in 
this  sample  of  oxide,  i.e.,  22.4/9.6,  must  be  equal  to  the 
ratio  between  the  formula-quantities  of  each  corresponding 
element  in  the  molecular  weight. 

Since  the  number  of  atomic  weights  of  each  element, 
grouped  in  its  formula-quantity,  is  here  unknown,  only 
the  algebraic  expression  ~FexOy  can  be  written  for  the 
complete  formula;  x  and  'y  representing,  of  course,  these 
unknown  integers.  Given,  then,  the  atomic  weights  of 
these  two  elements,  iron  (55.85)  and  oxygen  (16),  we  may 
easily  draw  up  the  ratio  between  the  formula-quantities  of 
each  in  the  compound.  This  ratio  involves  an  unknown 
factor  (an  integer)  in  each  term,  but  as  a  ratio  it  always 
must  be  equal  to  any  other  ratio  that  can  be  drawn  up 
from  data  on  the  weights  of  these  same  particular  sub- 
stances in  a  definite  weight  of  the  compound.  Thus  the 
ratio  Fez/Oj,.,  as  determined  in  the  analysis  of  the  oxide 
given  above,  is  represented  by  22.4/9.6,  and  consequently, 


DERIVATION   OF   CHEMICAL  FORMULAE  69 

if  the  analyses  are  correct,  these  two  ratios  must  be 

identical: 

Fe*       22.4         55.85  x  _  22 A 

Oy   " "   9.6  '  C        16  y    "  "   9.6  " 

The  exact  numerical  relation  between  x  and  y  may  be 
readily  calculated  by  bringing  the  numerical  values  to  one 
side  of  the  equation  (i.e.,  by  multiplying  the  equation 
through  by  16/55.85),  when  we  obtain 

z          22.4  X  16         358.4 

y     ~  9.6  X  55.85  "  536.4' 

a  result  likewise  obtained  through  the  simple  proportion: 

55.85  x  :  16  y  =  22A  :  9.6 
(55.85  X  9.6)  x  =  (16  X  22.4)  y 
x/y  =  358.4/536.2. 

In  one  equation  involving  two  unknown  terms  we  can 
only  expect  to  determine  the  simplest  ratio  between  them. 
The  ratio  358.4/536.4  when  reduced  by  division  of  each 
term  by  the  smaller  (358.4),  gives  us  1/1.49,  and  this  in 
turn,  through  multiplication  of  each  term  by  2,  is  brought 
very  close  to  2/3,  which,  as  the  ratio  between  the  smallest 
integers,  indicates  the  value  of  x  and  y  respectively; 
hence  the  formula  of  the  compound,  Fe2O3. 

Whether  the  molecular  formula  calls  for  this  simplest 
form  or  some  multiple  of  it  by  a  simple  integer  cannot  be 
determined  without  knowledge  of  the  molecular  weight 
of  the  compound. 

The  application  of  this  method  to  compounds  contain- 
ing three  or  more  elements  introduces,  of  course,  this 
corresponding  number  of  unknown  terms  (the  number 
of  atomic  weights  of  each  element  present),  and  lends 
itself  less  readily  to  a  simple  solution.  When,  however, 
two  or  more  groups  of  elements,  as  radicals,  etc.,  are 
present,  the  method  may  be  applied  to  the  determination 
of  the  number  of  each  of  the  two  groups  in  the  molecule. 


70  CHEMICAL   CALCULATIONS 

Example  31 .  —  What  is  the  formula  of  that  oxy-halogen  salt 
of  potassium,  (KCl)yOX)  13.9  grams  of  which  lost  6.4  grams  of 
oxygen  upon  heating,  and  gave  a  residue  of  7.5  grams  (13.9-6.4) 
of  potassium  chloride,  KC1? 

The  ratio  between  the  weight  of  oxygen  driven  off  and 
the  residue  of  potassium  chloride  left  is  6.4/7.5,  which,  of 
course,  is  equal  to  the  ratio  between  the  formula-quantity 
of  each: 

Ox       =  6.4    or  (16)  x          =  6.4 

(KCl)y      7.5'          (39.1+35.46)  y      7.5' 
or 

£       (6.4)  (74.56)       476.2 
y  =       (7.5)  (16)  120  ' 

From  this  we  obtain  x/y  —  3.97/1  or,  by  simplest  integers, 
#  =  4  and  y=  I.  The  formula  of  the  compound  (KC1)VOZ 
now  becomes  (KCi)104,  i.e.,  KC1O4. 

Formulae  Containing  Water  of  Crystallization.  —  Again 
we  have  a  good  illustration  of  examples  of  this  kind  in 
the  consideration  of  those  salts  which  contain  water  of 
crystallization,  —  definite  in  amount  under  certain  defi- 
nite and  fixed  conditions. 

Example  82.  — 7.15  grams  of  crystallized  sodium  carbonate, 
NaaCOg .  x  H2O,  lost  all  of  its  water  of  crystallization  when 
gently  heated.  The  weight  of  the  residue  was  2.655  grams. 
What  is  the  formula  of  the  crystallized  salt? 

Here  7.15  —  2.655  =  4.495  grams,  or  the  weight  of 
water  in  this  weight  of  salt.  The  ratio  between  the  weight 
of  water  actually  associated  and  the  weight  of  anhydrous 
salt  (that  portion  of  the  crystallized  salt  left  after  the 
elimination  of  the  water)  is  4.495/2.655.  This  ratio  there- 
fore must  be  equal  to  the  ratio  between  the  formula-quan- 
tity of  water  and  the  formula-quantity  of  the  anhydrous 
salt  in  the  molecular  weight  of  the  crystallized  salt.  In 
that  the  water  of  crystallization  is  usually  determined 


DERIVATION    OF  CHEMICAL   FOKMULuE  71 

with  reference  to  one  molecule  of  the  anhydrous  salt,  one 
of  the  unknown  terms  in  the  ratio  between  the  formula- 
quantities  drops  out  (i.e.,  equals  unity)  and  we  have  only 
the  determination  of  the  unknown  number  of  molecular 
weights  (x)  of  water  present.  Thus: 

s(H2O)       4.495 
Na,CO3  ~  2.655 ' 

By  substituting  the  atomic  weights  indicated,  we  obtain 

x  (18.02)  =  4.495 
106          2.655 ' 

which,  when  solved  for  x,  gives  the  value  10  as  the  number 
of  molecules  of  water  associated  with  one  molecule  of  the 
anhydrous  salt  to  form  a  molecule  of  crystallized  sodium 
carbonate,  Na2C03 .  10  H20. 

The  calculation  may  be  conducted  by  simple  proportion 
as  follows: 

Na2CO3 :  x  (H20)    =  wt.  anhydrous  salt :  wt.  H2O. 
106  :  x  (18.02)  =  2.655  :  4.495. 

The  expression  x  (18.02)  is  found  equal  to  179.5,  hence  the 
integral  value  for  x  will  be  10  and  accord  with  the  formula 
above. 

Formulae  from  the  Relation  between  Formula-Quan- 
tities and  Their  Corresponding  Weights.  —  This  constancy 
in  the  ratios  between  the  quantities  of  all  the  elements 
either  singly  or  collectively  in  a  chemical  compound  was 
comprehended,  of  course,  in  the  Law  of  Definite  Pro- 
portions; and  the  proportions  in  which  the  elements  enter 
into  chemical  combination  are  seen,  accordingly,  to  be 
functions  of  the  corresponding  atomic  weights.  In  any 
known  amount  of  substance  there  is  always  a  definite 
ratio  between  the  actual  weight  of  some  one  element,  or 
group  of  elements,  present  and  the  corresponding  formula- 


72  CHEMICAL   CALCULATIONS 

quantity.  Furthermore,  this  ratio  is  always  identical  with 
every  other  ratio  between  the  weight  of  any  other  element 
in  this  same  sample  and  its  corresponding  formula-quan- 
tity.* The  ratio  can  be  unity  only  when  we  are  dealing 
with  a  gram-molecular  weight  of  the  substance.  Thus, 
in  18  grams  of  water,  the  weight  of  oxygen  (16  grams) 
bears  the  ratio  of  unity  to  the  formula-quantity  of  this 
element  in  the  molecular  weight,  and  also  the  same  ratio, 
unity,  exists  between  the  weight  of  hydrogen  (2  grams) 
and  its  formula-quantity. 

As  is  often  the  case,  no  one  correct  formula-quantity 
is  at  hand.  Under  such  circumstances  it  is  only  possible 
to  refer  any  and  all  of  the  known  weights  of  the  elements 
in  a  given  sample  of  compound  directly  to  their  corre- 
sponding atomic  weights.  In  order  to  maintain  through- 
out the  same  constant  ratio  between  these  values  — 
actual  weights  and  atomic  weights  —  it  is  necessary  to 
select  the  ratio  between  the  actual  weight  of  some  one 
element  in  the  sample  and  the  atomic  weight  correspond- 
ing thereto  as  the  standard  ratio.  By  means  of  this  ratio 
the  actual  weight  of  any  other  element  present  in  the 
sample  may  be  brought  over  immediately  (simple  pro- 
portion) to  a  value  which  holds  the  same  relation  to  its 
true  formula-quantity  as  the  single  atomic  weight  of  the 
first  element  will  be  found  to  have  toward  its  formula- 

*  The  equivalence  in  the  ratios  between  any  two  formula-quantities, 
as  x  and  y,  and  the  corresponding  weights,  as  w^  and  w2,  which  repre- 
sent them  respectively  in  any  given  sample,  is  usually  expressed  as 
x/y  ==  wjwv  By  multiplying  through  by  y/wl  we  immediately 
obtain  the  expression  x/w^  =  y/wz  for  this  equivalence  in  the  ratios 
between  each  formula-quantity  and  the  weight  representing  it  in  some 
sample.  The  two  expressions  take  the  following  forms  through 
simple  proportion:  x  :  y  —  wt :  w2  and  x  :  wl  =  y  :  wv  each  of  which 
follows  algebraically  from  the  other,  i.e.,  the  ratio  between  the  first 
and  second  terms  of  a  proportion,  when  equal  to  the  ratio  between  the 
third  and  fourth  terms,  signifies  also  this  equivalence  in  the  ratios 
between  the  first  and  third  terms  and  the  second  and  fourth  terms. 


DERIVATION   OP   CHEMICAL   FORMULAE  73 

quantity.  In  this  latter  case,  with  the  formula-quantity 
always  a  multiple  of  the  atomic  weight  by  a  small  integer, 
the  process  of  determining  the  true  formula-quantity  is 
comparatively  a  simple  one.  The  integer  selected,  how- 
ever, must  also  bring  each  and  all  of  the  other  values, 
similarly  found,  to  multiples  of  their  individual  atomic 
weights.  In  other  words,  we  determine  the  least  common 
multiple  for  the  entire  range  of  values,  such  that  each  in 
order  will  be  raised  to  a  multiple,  by  unity  or  a  small 
integer,  of  its  atomic  weight.  Finally,  the  quotients  of 
these  so  derived  formula-quantities  by  the  corresponding 
atomic  weights  will  give  the  units  of  each  element  neces- 
sary in  the  simplest  formula;  one  which  can  be  raised 
later  to  any  desired  multiple,  depending,  of  course,  upon 
the  molecular  weight  in  question. 

Example  S3.  —  Derive  the  formula  of  iron  oxide  from  the  data 
in  Example  30  :  22.4  grams  of  iron  gave  32  grams  of  oxide. 

The  weight  of  oxygen  entering  into  the  compound  is 
9.6  grams.  Accordingly,  the  ratio  9.6/ 16,  between  weight 
of  oxygen  present  in  the  sample  and  the  atomic  weight 
of  this  element,  may  be  taken  as  the  standard  ratio. 
By  means  of  this  ratio  the  weight  of  iron  present  is  now 
brought  over  to  a  value  always  equivalent  to  oxygen  as 
16  in  this  compound.*  Thus,  9.6  :  16  =  22.4  :x,  where 

*  In  the  combinations  between  hydrogen  and  oxygen  no  com- 
pound has  been  found  to  contain  more  than  two  atomic  weights  of 
hydrogen  with  one  of  oxygen  in  a  single  molecular  weight.  Con- 
sequently the  lowest  weight  of  oxygen  corresponding  to  the  lowest 
weight  of  hydrogen  (1.008,  its  atomic  weight)  will  be  one-half  of 
16,  or  8.  This  weight  of  oxygen,  the  smallest  equivalent  to  one 
atomic  weight  of  hydrogen,  is  often  used  as  a  basis  for  the  conver- 
sion of  known  weights  or  percentages  of  other  elements  (associated 
with  oxygen)  to  the  scale  of  atomic  weights.  Though  it  brings  into 
the  calculations  numbers  somewhat  smaller  (by  one-half)  than  other- 
wise obtained  through  the  use  of  the  actual  atomic  weight,  16,  the 
process  of  determining  the  least  common  multiple  is  in  no  way 
simplified. 


74 


CHEMICAL   CALCULATIONS 


this  value  for  iron  is  found  to  be  37.2.     By  referring  these 
values  now  to  the  respective  atomic  weights  we  have: 


Relative 

Value 

Sub- 
stance. 

amount  in 
known 
weight  of 

Value  con- 
structed 
upon  oxy- 

Multiple 

raised  to 
multiple  of 
correspond- 

Atomic 
weight. 

Unit 
value. 

compound. 

gen  as  16. 

. 

ing  at.  wt. 

Iron.  .  . 

22.4 

37.2 

//>/ 

111.6 

55.85 

2 

Oxygen 

9.6 

16 

^ 

48 

16 

3 

From  this  the  formula  of  the  compound  is  found  to 
Formulae  from  the  Relation  between  Formula-Quantity 
and  Percentage  Composition.  —  In  extending  this  method 
to  compounds  which  contain  a  number  of  elements,  it 
serves  equally  well  if  we  employ,  in  place  of  actual  weights, 
the  percentage  amounts  of  the  elements,  or  groups  of  ele- 
ments, present.  These  values,  of  course,  are  as  definitely 
related  to  each  other  as  any  other  fractions  of  the  total 
weight  of  a  substance.  The  ratio  between  the  percentage 
of  any  one  element  and  its  atomic  weight  suffices,  through 
simple  proportion,  to  bring  over  each  of  the  other  per- 
centages to  the  corresponding  values  for  these  elements 
in  the  compound. 

Example  34-  —  Derive  the  formula  of  sodium  sulphate  from 
the  data  given  in  Example  27. 


Sub- 
stance. 

Percent- 
age. 

Value 
con- 
structed 
upon  oxy- 

Multiple 

Value 
raised  to 
multiple  of 
correspond- 

Atomic 
weight. 

Unit 
value. 

gen  as  16. 

~~~ 

' 

fit; 

ing  at.  wt. 

Sodium.  .  . 
Sulphur..  . 
Oxygen.  .  . 

32.32 
22.44 
45.24 

11.43 
7.94, 
16 

45.72 
31.76 
64 

23 
32.07 
16 

1.99 
0.99 
4 

100.00 

DERIVATION   OF   CHEMICAL  FORMULAE  75 

The  value  of  oxygen  is  made  equal  to  the  atomic  weight 
of  this  element,  and  the  ratio  45.06  :  16  is  used  to  reduce 
all  of  the  other  values  proportionately,  e.g.: 

with  sulphur       45.24  :  16  =  22.44  :  x,  or  x  =  7.94  and 
with  sodium        45.24  :  16  =  32.32  :  x,  or  x  =  11.43. 

These  final  values  represent  correctly  the  relative  amounts 
of  each  element.  One,  oxygen,  was  made  to  coincide 
with  its  corresponding  atomic  weight,  hence  each  of  the 
other  values  will  represent  a  weight  of  that  respective  ele 
ment  in  this  compound  equivalent  to  oxygen  as  16;  and, 
as  this  is  the  unit  value  for  oxygen,  the  entire  range  of 
values  can  be  raised  only  through  multiplication  by  simple 
integers  until  the  lowest  possible  formula-quantity  for 
each  is  obtained.  The  integer  4,  found  by  trial,  here 
raises  all  to  values  that  are  multiples  of  the  corresponding 
atomic  weights.  These  formula-quantities  when  divided 
by  the  respective  atomic  weights  give  at  once  the  number 
of  atomic  weights  of  each  element  in  the  molecule,  hence 
the  formula  Na2SO4. 

The  Derivation  of  Formulae  of  a  Known  Type  from  a 
Single  Analytical  Value.  —  If  it  is  known  to  what  general 
class  (oxides,  chlorides,  sulphates,  etc.)  a  particular  sub- 
stance of  simple  type  (not  involving  replacements)  belongs, 
its  complete  analysis  is  not  necessary  for  the  determination 
of  the  formula.  For  instance,  the  percentage  of  copper 
in  a  sample  known  to  be  a  copper  chloride  will  be  suffi- 
cient to  this  end.  This  follows  by  reason  of  the  fact  that 
the  percentage  amount  of  any  element  found  comes  from 
the  ratio  between  its  formula-quantity  in  the  molecule 
and  the  total  molecular  weight,  —  a  ratio  represented, 
we  shall  say,  by  the  expression,  — mol.  wt./x  (at.  wt.). 
Upon  analysis  certain  definite  numerical  values  fall  to 
this  ratio;  in  percentages,  for  example,  we  have  the 


76  CHEMICAL  CALCULATIONS 

ratio  100 /per  cent  element  found.  We  thus  form  the 
equation: 

100  _=  mol.  wt. 

%  element  found  ~~  x  (at.  wt.  element) 

Example  35.  —  A  sample  of  a  chloride  of  copper  gave  upon 
analysis  47.9  per  cent  copper.  What  is  the  formula  of  the  com- 
pound? 

The  ratio  between  the  known  and  theoretical  values 
now  becomes 

100/47.9    =  mol.  wt./z  (63.57), 
or  100  :  47.9  =  mol.  wt.  :  x  (63.57). 

As  x  is  always  a  small  integer  it  may  be  neglected  for 
the  first  calculation:  100  :  47.9  =  mol.  wt.  :  63.57.  From 
this  we  obtain  a  value  132.71,  as  the  molecular  weight, 
in  which  one  atomic  weight  of  copper  must  be  present 
to  the  extent  of  47.9  per  cent.  The  difference,  therefore, 
or  69.14  (i.e.,  132.71  -  63.57  =  69.14),  must  represent  the 
atomic  weights  of  chlorine  associated  with  this  one  atomic 
weight  of  copper.  The  quotient  of  69.14  by  35.46,  the 
atomic  weight  of  chlorine,  gives  1.95,  or  practically  2,  and 
we  at  once  draw  up  the  formula  CuCl2  (63.57  +  2  (35.46) 
=  134.49),  with  the  molecular  weight  of  which  the  an- 
alytical data  are  well  in  agreement;  hence  the  correct 
formula. 

If  there  had  been  obtained  some  fractional  quantity  as  a 
quotient  (in  place  of  2  above),  this  quantity,  together  with 
the  value  for  copper  as  unity,  may  be  raised  by  multiplica- 
tion to  integral  values  which  indicate  the  probable  formula. 
This,  of  course,  is  identical  with  multiplying  the  molecular 
weight  found  by  x  (the  smallest  integrals  in  their  order) 
until  a  value  is  obtained  that  can  be  made  up  of  atomic 


DERIVATION   OF   CHEMICAL   FORMULAE  77 

weights  without  fractional  parts.  This  try-out  of  a  for- 
mula is  nothing  more  than  the  reverse  of  the  method 
outlined  in  Example  23,  wherein  was  shown  the  means  of 
determining  the  amount  of  any  salt  present  in  any  one  of 
its  samples  when  given  the  percentage  of  some  one  element 
in  the  salt.  In  place  of  the  molecular  weight  there  given 
we  have  here  only  to  set  down,  in  order,  the  trial  values 
for  this  molecular  weight  which  correspond  to  the  possible 
formula.  Thus: 

(  Percentage  amount )  (  _  _  )  _  (  Formula-quantity )  ( Molecular  weight ) 
j  of  copper  )  *  j  )  i  copper  }  ''  \  calculated.  ) 

47.9  :     x       =  63.57  : 

When  the  value  calculated  for  x,  from  any  trial  molecular 
weight,  approaches  closely  to  100  we  are  assured  of  the 
correct  value  for  this  molecular  weight  and  consequently 
of  the  correct  formula. 


PROBLEMS. 

99.  What  is  the  formula  of  that  substance  which  gave,  by 
analysis,  26.9  per  cent  sodium,  16.58  per  cent  nitrogen,  56.52 
per  cent  oxygen?  Ans.   NaNO3. 

100.  What  is  the  formula  of  that  substance  which  gave,  by 
analysis,  26.6  per  cent  potassium,  35.22  per  cent  chromium, 
38.18  per  cent  oxygen?  Ans.   K2Cr2O7. 

101.  What  is  the  formula  of  that  substance  which  gave,  by 
analysis,  24.65  per  cent  potassium,  34.85  per  cent  manganese, 
40.5  per  cent  oxygen?  Ans.   KMnO4. 

102.  Derive  the  formula  of  that  substance  with  the  observed 
relative  density  0.8853(0=  1)  and,  by  analysis,  the  composition: 
85.41  per  cent  carbon  and  14.64  per  cent  hydrogen.  Ans.  C2H4. 

103.  Derive  the  formula  of  that  substance  with  the  observed 
density  1.189  and,  by  analysis,  the'  composition:  92.1  per  cent 
carbon,  7.85  per  cent  hydrogen.  Ans.   CJEL*. 


78  CHEMICAL   CALCULATIONS 

104.  Derive  the  formula  of  that  compound  of  hydrogen  and 
oxygen  which  gave,  by  analysis,  5.93  per  cent  hydrogen.     A 
determination  of  its  molecular  weight  gave  the  value  31.8. 

Am.  H202. 

105.  Derive  the  formulae  of  that  oxide  of  nitrogen  which 
gave,  by  analysis,  30.4  per  cent  nitrogen.     In  the  solid  state  it 
was  found  to  have  a  molecular  weight  of  92.4,  whereas  the 
actual  density  of  its  vapor  above  140°  was  only  2.013. 

A         (  Solid,  N204. 
i  Gas,  N02.  ? 

106.  Derive  the  formula  of  that  acid  which  gave,  by  analysis, 
26.5  per  cent  carbon,  2.2  per  cent  hydrogen  and  the  rest  oxygen. 
The  molecular  weight  was  determined  approximately  as  90.4. 

Ans.    H2C204. 

107.  Derive  the  formula  of  the  mineral  chalcopyrite,  a  speci- 
men of  which  gave,  by  analysis,  the  following  percentage  compo- 
sition: 34.40  per  cent  copper,  30.47  per  cent  iron  and  35.87 
per  cent  sulphur.  Ans.   CuFeS,. 

108.  Derive  the  formula  of  the  mineral  dolomite  which  gave, 
by  analysis,  the  following  percentage  composition  : 


C02  47.67  per  cent 

100.27 

Ans.    (CaO,MgO)(C02) 
or     (Ca,Mg)  C03. 

109.   By  analysis  a  specimen  of  melanterite  gave  the  fol- 
lowing: 

FeO         20.37  per  cent  )  . 

TIT  ^  A  ™  i  isomorphous 

MgO          4.60  per  cent  ) 

SO3          29.80  per  cent 
H20         45.07  per  cent 


99.84 
Derive  the  formula. 


Ans.    (FeO,MgO)(S03)  .  7  H30 
or  (Fe,Mg)S04.7HA 


DERIVATION    OF   CHEMICAL   FORMULAE 


79 


110.  By  analysis  a  specimen  of  xenotime  gave  the  follow- 


ing: 


PA 

Y203 
Ce203 
Fe203 


32.45  per  cent 

54.13  per  cent 

11.03  per  cent 

2.06  per  cent 


isomorphous 


99.67 
Derive  the  formula. 

Arts.    (Y203,Ce203,Fe203)  .  (P2O6) 
or  ([Y,Ce,Fe]203)  (P2O5) 

111.  By  analysis  a  specimen  of  columbite  gave  the  following: 

Nb205  47.05  per  cent 

Ta206  34.04  per  cent 

Sn02  0.30  per  cent 

FeO  11.15  per  cent 
MnO 


7.80  per  cent  , 


isomorphous 
isomorphous 


100.34 
Derive  the  formula. 


Ans.    ([F< 


,Ta]A). 


,0.7*0.     ^[j? Cjivj-iijOujvy,;  .  ^LJ.NU, xaj2v 
112.   By  analysis  a  specimen  of  garnet  gave  the  following: 

c:rv  39.09  per  cent 

23.05  per  cent 

0.53  per  cent 

0.11  per  cent 

0.35  per  cent 

1.01  per  cent 
35.75  per  cent 

0.15  per  cent     (neglected) 


2 

A12O3 
Fe2O3 
FeO 
MnO 
MgO 
CaO 
H20 


isomorphous 
isomorphous 


100.04 
Derive  the  formula. 

Ans.   3  ([Ca,Mg,Mn,Fe]0)  .  ([Fe,Al]2O3)  .  3  (Si02) 
General  type,  3  RO  .  R2O3 .  3  SiO2 
or  R"3R"'2(SiO4)3. 

113.   Derive  the  formula  of  the  oxide  produced  when  6.87 
grams  of  barium  unite  with  1.6  grams  of  oxygen. 

Ans.    Ba03. 


80  CHEMICAL   CALCULATIONS 

114.  Derive  the  formula  of  the  oxide  formed  in  the  com- 
bustion of  2.61  grams  of  aluminium  with  oxygen  to  a  final 
weight  of  5.01  grams.  Ans.    A1203. 

115.  Derive  the  formula  of  the  oxide  produced  by  the  com- 
bustion of  43.45  grams  of  lead  with  4.48  grams  of  oxygen. 

Ans.   Pb3O4. 

116.  Derive  the  formula  of  the  oxide  produced  by  the  burn- 
ing of  2.5  grams  of  phosphorus  in  oxygen  to  a  final  weight  of 
5.7  grams.  Ans.    P2O5. 

117.  Derive  the  formula  of  the  nitrate,  19.7  grams  of  which 
were  prepared  from  10.4  grams  of  bismuth. 

Ans.  Bi(N03)3. 

118.  Derive  the  formula  of  the  nitrate,  17  grams  of  which 
gave  a  residue  of  13.8  grams  of  sodium  nitrite,  NaN02,  upon 
heating.  Ans.  NaN03. 

119.  Derive  the  formula  of  that  chlorate,  4.165  grams  of 
which  lost  1.3  grams  of  oxygen  upon  heating  and  gave  a  resi- 
due of  barium  chloride,  BaCl2.  Ans.  Ba(C103)2. 

120.  Derive  the  formula  of  mercuric  cyanide,  5.4  grams  of 
which  lost  1.1  grams  of  cyanogen  upon  heating. 

Ans.  HgC2N2. 

121.  Derive  the  formula  of  the  double  salt  of  ammonium 
sulphate  and  copper  sulphate,  4.12  grams  of  which  lost  1.81 
grams  of  ammonium  sulphate  upon  heating. 

Ans.  (NH4)2S04 .  CuS04. 

122.  Derive  the  formula  of  crystallized  sodium  sulphate,  8.16 
grams  of  which  lost  4.51  grams  of  water  upon  dehydration. 

Ans.  Na2S04 .  10  H2O. 

123.  Derive  the  formula  of  crystallized   copper  sulphate, 
7.84  grams  of  which  lost  2.79  grams  of  water  upon  dehydration. 

Ans.  CuS04 .  5  H20. 

124.  Derive  the  formula  of  crystallized  aluminium  sulphate, 
9.54  grams  of  which  lost  4.61  grams  of  water  upon  dehydration. 

Ans.  A12(S04)3 .  18  H20. 

125.  Derive  the  formula  of  aluminium  hydroxide,  4.75  grams 
of  which  lost  1.64  grams  of  water  and  left  a  residue  of  A1203. 

Ans.  Al(OH),. 


DERIVATION    OF   CHEMICAL   FORMULAE  81 

126.  Derive  the  formula  of  that  nitrate  which  gave,  by 
analysis,  62.45  per  cent  lead,  8.68  per  cent  nitrogen,  28.85 
per  cent  oxygen.  Ans.  Pb(N03)2. 

137.  Derive  the  formula  of  that  substance  which  gave,  by 
analysis,  52.02  per  cent  carbon,  13.2  per  cent  hydrogen,  34.68 
per  cent  oxygen.  Ans.  C2H60. 

138.  Derive  the  formula  of  that  acetate  which  gave,  by  analy- 
sis, 63.61  per  cent  lead,  14.62  per  cent  carbon,  1.98  per  cent 
hydrogen,  19.79  per  cent  oxygen.  Ans.  Pb(C2H302)2. 

129.  A  salt  of  mercury  known  to  be  a  chloride  analyzed  for 
26.15  per  cent  chlorine.     What  is  the  formula?      Ans.  HgCl2. 

130.  A  salt  known  to  be  a  nitrate  analyzed  for  62.4  per  cent 
lead.    What  is  the  formula?  Ans.  Pb(N03)2. 

131.  An  oxide  of  iron  gave,  by  analysis,  69.80  per  cent  iron. 
What  is  the  formula?  Ans.  Fe2O3. 

132.  An  oxide  of  barium  gave,  by  analysis,  81  per  cent 
barium.    What  is  the  formula?  Ans.  Ba02. 

133.  Derive  the  formula  of  the  crystallized  salt  which,  by 
analysis,  gave  the  percentage  composition:  19.98  per  cent  iron, 
11.47  per  cent  sulphur,  5.24  per  cent  hydrogen,  63.31  per  cent 
oxygen,     10  grams  of  this  crystallized  salt  lost  4.5  grams  of 
water  upon  dehydration.  Ans.  FeSO4 .  7  H20. 

134.  The  percentage  composition  of  a  certain  salt  is:  15.6 
per  cent  chromium,  14.41  per  cent  sulphur,  4.79  per  cent  hydro- 
gen,  65.2  per  cent  oxygen.     10  grams  of  the  crystallized  salt 
lost  4  grams  of  water  upon  dehydration  and  gave  a  residue  of 
Cr2(S04)3.    What  is  the  formula?        Ans.  Cr2(SO4)3 .  15  H2O 

135.  The  percentage  composition  of  a  certain  salt  is:  27.51 
per  cent  calcium,  22.15  per  cent  sulphur,  1.02  per  cent  hydrogen, 
49.32  per  cent  oxygen.     10  grams  of  this  crystallized  salt  lost 
0.6  gram  of  water  upon  dehydration.    What  is  the  formula? 

Ana.  (CaS04)2 .  H2O. 

136.  2.5  grams  of  a  crystallized  salt  known  to  be  a  sulphate 
of  iron,  and  containing  20  per  cent  of  iron,  lost  1.13  grams  of 
water  upon  dehydration.    What  is  the  formula? 

Ans.  FeS04 .  7  HaO. 


CHAPTER  IX. 

CALCULATIONS   DEPENDING   UPON    CHEMICAL 
EQUATIONS. 

IN  calculations  which  depend  upon  chemical  reactions, 
the  equations  representing  these  reactions  must  first  be 
constructed.  In  all  chemical  equations  the  number  of 
atomic  weights  of  any  one  element  concerned  remains  a 
constant;  the  relative  amount  of  each  element,  therefore, 
will  be  alike  for  both  sides  of  the  equation.  Naturally 
the  valence  of  each  element  in  the  reaction  under  consid- 
eration must  be  known,  as  upon  these  factors  the  balancing 
of  equations  is  dependent.  This  valence  or  measure  by 
which  each  atom  of  any  element  can  enter  into  combina- 
tion determines,  accordingly,  the  number  of  other  atoms  or 
groups  of  atoms  necessary  for  consideration  in  any  reac- 
tion. When,  however,  an  element  undergoes  a  change  of 
valence  the  arrangement  of  these  atom  groupings  must  be 
made  to  accord  with  this  change  (cf.  Chap.  XII).  The 
representation  of  all  substances  in  the  molecular  form  — 
constituting  here  a  molecular  equation  —  has  the  great 
advantage  of  indicating  the  corresponding  volume  rela- 
tions between  the  gaseous  substances  by  reason  of  the  like 
molecular  volumes  of  all  substances  in  the  state  of  vapor. 

Reaction-Quantities.  —  The  action  of  metallic  sodium 
upon  water  is  shown  in  the  following  equation : 

2  Na        +  2  H2O       =  2  NaOH  +  H2 
2  mol.      +  2  mol.       =  2  mol.      +  1  mol. 

Relative          j  2  (23.00)  +  2  (18.02)  =  2  (40.01)  +  2.02 

parts 

by  weight.       I  46.00        +  36.04        =  80.02        +  2.02 

82.04        =  82.04 

8* 


CHEMICAL   EQUATIONS  83 

In  this  simple  equation  each  of  the  quantities  repre- 
sented is  definite  and  bears  that  relation  to  every  other 
quantity  as  is  indicated  by  the  corresponding  group  of 
atomic  weights  present  in  each.  These  quantities  may 
be  considered  as  the  reaction-quantities,  —  atomic  or 
molecular  quantities  definite  for  any  given  reaction  and 
always  proportional  to  each  other,  such  that,  when  in- 
volved together,  the  ratio  between  them  will  be  exactly 
equal  to  the  ratio  between  the  actual  weights,  of  whatever 
denomination,  which  may  represent  them.  This  propor- 
tionality, relative,  of  course,  to  the  construction  of  the 
equation  itself,  follows  naturally  from  the  Law  of  Definite 
Proportions.  The  reaction  equation  only  indicates  the 
apportionment  of  the  various  atomic  or  molecular  group- 
ings for  the  several  compounds  possible  under  the  observed 
conditions.  The  formula-quantities  going  to  make  up  any 
number  of  molecules  of  a  compound  indicated,  still  bear  a 
definite  ratio  to  every  other  formula-quantity,  with  the 
result,  of  course,  that  the  weight  which  represents  any  of 
these  reaction-quantities  in  a  given  equation  must  also 
bear  a  similar  and  definite  proportion  to  each  other,  no 
matter  whether  they  are  upon  the  same  or  opposite  sides 
of  the  equation.  The  reaction-quantity  of  sodium  in  the 
equation  above  is  represented  by  two  molecular  weights; 
the  reaction-quantity  of  sodium  hydroxide,  corresponding 
to  this  value  for  sodium,  is  also  represented  by  two  molec- 
ular weights.  As  both  involve  the  same  quantity  of 
sodium  they  may  be  regarded  also  as  equivalent 
quantities.  The  ratio  between  them,  2  Na/2  NaOH,  is 
more  simply  expressed  in  the  unimoleeular  form, 
Na/NaOH. 

The  reaction-quantity  of  hydrogen  directly  proportional 
to  the  reaction-quantity  of  sodium  in  this  equation  is  rep- 
resented by  one  molecular  weight,  or  H2.  The  ratio 


84  CHEMICAL   CALCULATIONS 

2  Na/H2  is  definite,  therefore,  for  all  possible  values  for 
these  substances  in  this  reaction. 

Example  86.  —  What  weight  of  sodium  hydroxide  and  of 
hydrogen  can  be  procured  by  the  action  of  50  grams  of  sodium 
upon  water? 

From  the  equation  just  studied,  the  ratio  2  Na/2  NaOH, 
or  Na/NaOH,  i.e.,  23/40.01,  is  constant  for  all  proportional 
amounts  of  these  two  substances.  Therefore  the  ratio 
50 / x,  between  this  known  weight  of  the  metal  (50  grams) 
and  its  proportional  value  in  hydroxide  (x),  must  be 
equal  to  the  ratio  above  (the  related  terms  in  these  ratios 
are  of  course  placed  similarly),  and  we  shall  have 

23/40.01  =  50/s. 

By  simple  proportion  this  may  be  expressed  as 
23  :  40.01  =  50  :  x. 

From  these  expressions  the  value  of  x  (the  weight  of 
sodium  hydroxide)  is  found  to  be  86.98  grams. 

In  an  exactly  similar  manner  the  ratio  between  the 
reaction-quantities  of  sodium  and  hydrogen,  2  Na/H2,  or 
Na/H,  or  23/1.01  may  be  made  equal  to  the  ratio  between 
the  known  or  unknown  weights  of  these  substances  con- 
cerned in  the  reaction.  Between  the  weight  of  sodium 
(50  grams)  and  its  unknown  proportional  weight  of 
hydrogen  (x)  we  have  the  ratio  50  /x;  this  is  therefore 
immediately  referred  to  the  ratio  between  the  correspond- 
ing proportional  quantities  above,  Na/H  or  23/1.01. 
Thus,  23/1.01  =  5Q/x,  or  23  :  1.01  =  50  :  x,  from  which 
the  value  for  x  is  found  to  be  2.2  grams. 

In  general  we  may  state  that  when  any  substance  is 
concerned  in  a  chemical  reaction,  the  amount  of  any  other 
substance,  resulting  directly  or  indirectly  through  this 
reaction,  bears  to  the  former  a  definite  ratio  and  one  always 


CHEMICAL   EQUATIONS  85 

equal  to  the  ratio  between  the  corresponding  reaction- 
quantities  of  the  substances  in  the  equation. 

Example  37.  —  What  weight  of  magnesium  will  be  required 
for  the  liberation  of  10  grams  of  hydrogen  from  water  or  an  acid? 

We  may  first  consider  the  decomposition  of  water  by 
magnesium  as  shown  in  the  following: 

Mg        +  H20    =  MgO    +  H2 
1  mol.  +  1  mol.=  1  mol.  +  1  mol. 
Parts  by  weight.  {24.32    +18.02  =  40.32  +  2.02 

and  again,  the  reaction  of  magnesium  upon  an  acid  such 

as  hydrochloric  acid: 

Mg      +2HC1       =MgCl2+H2 
1  mol.  +  2  mol.       =  1  mol.  +  1  mol. 

Parts  by  weight.  (24.32  +  2  (36.47)  =  95.24  +  2.02 

In  each  case  we  find  that  the  molecule  of  magnesium  — 
containing  one  atomic  weight  —  displaces,  and  is  equi- 
valent to,  one  molecule  of  hydrogen  —  containing  two 
atomic  weights.  The  quantities  Mg  and  H2  are  there- 
fore directly  proportional  to  each  other  in  both  equations, 
and  the  constant  ratio  Mg/H2,  or  24.32/2.02,  must  be  equal 
to  the  ratio  between  the  corresponding  weights  of  these 
elements  here  concerned,  i.e.,  z/10.  From  the  equation 
24.32/2.02  =  xl  10,  or  the  simple  proportion  24.32  :  2.02 
=  x  :  10,  we  calculate  the  value  of  x  to  be  120.4  grams 
(magnesium). 

In  the  more  complicated  reactions  the  relations  between 
the  several  quantities  on  the  two  sides  of  an  equation  are 
often  difficult  to  ascertain.  To  say  nothing  of  reversible 
actions  and  the  possibilities  for  dependent  or  secondary 
reactions  to  take  place  between  certain  of  the  quantities, 
we  have  assumed  and  must  continue  to  assume  that  the 
reaction  in  question  takes  place  along  the  lines  indicated 


86  CHEMICAL   CALCULATIONS 

in  that  equation  best  substantiated  by  the  facts  under  the 
observed  conditions. 

Example  88.  —  What  weight  of  sulphur  dioxide,  S02,  can  be 
obtained  by  the  action  of  10  grams  of  copper  upon  concentrated 
sulphuric  acid? 

Copper  acts  upon  sulphuric  acid  (cone.),  to  give  sulphur 
dioxide,  water  and  copper  sulphate.  Notwithstanding 
the  slight  amount  of  cuprous  sulphide  formed  here  through 
a  secondary  reaction,  we  shall  represent  the  action  by  a 
single  equation: 

Cu      +2  H2S04  =  CuS04  +  2  H20  +  S02 
63.57  64.07 

The  reaction-quantities,  Cu  and  S02,  are  directly  pro- 
portional; the  ratio  between  them,  Cu/S02,  or  63.57/64.07, 
is  constant  for  all  amounts  here  involved,  and  conse- 
quently equal  to  the  ratio,  10  fx,  representing  these  sub- 
stances respectively  in  this  example.  From  the  equation 
63.57/64,07  =  W/x  the  value  of  x  is  calculated  as  10.08, 
the  weight  in  grams  of  sulphur  dioxide. 

Calculation  of  Volume  Relations  Introduced  by  Chem- 
ical Equations.  —  In  the  consideration  of  gaseous  sub- 
stances formed  in  these  reactions,  their  gram-molecular 
quantities  may  be  referred  at  once  to  the  gram-molec- 
ular volumes  as  a  basis  (cf.  Ex.  16).  For  example,  to 
recall  the  illustrations  at  the  beginning  of  this  chapter, 
two  gram-molecular  weights  of  sodium  displace  one  gram- 
molecular  weight  of  hydrogen,  which  under  standard 
conditions  occupies  22,400  c.c.  Whatever  be  the  weight 
of  hydrogen  evolved,  the  ratio  between  the  G.M.W.  and 
this  weight  will  be  identical  with  that  between  the  G.M.V. 
and  the  volume  of  hydrogen  corresponding  to  the  known 
weight  above  (cf.  Ex.  15). 


CHEMICAL   EQUATIONS  87 

Example  39.  —  Calculate  the  volume  of  hydrogen,  at  10°  and 
750  mm.,  that  will  be  evolved  by  the  action  of  100  grams  of  zinc 
upon  an  acid. 

The  nature  or  strength  of  the  acid  here  is  of  no  concern, 
for,  so  long  as  the  zinc  is  used  up,  the  proportional  amount 
of  hydrogen  must  be  displaced,  and  from  this  weight  of 
hydrogen  displaced  its  volume  may  be  easily  calculated. 
With  sulphuric  acid  the  reaction  may  be  represented  as 
follows : 

Zn      +  H2SO4  =  ZnSO4  +  H2 
65.37  +  98.09     =  161.44  +  2.02. 

The  reaction-quantities  Zn  and  H2  establish  the  ratio 
Zn/H2,  or  65.37/2.02,  as  constant  for  all  proportional 
amounts  of  these  elements;  hence  the  equation  65.37/2.02 
=  100/z.  From  this  the  value  of  x  is  calculated  as  3.09 
grams. 

A  gram-molecular  weight  of  hydrogen,  2.02,  occupies 
22,400  c.c.  at  standard  conditions.  As  previously  stated 
the  volume  occupied  by  any  other  weight  of  the  gas  will 
stand  in  the  same  proportion  to  this  gram-molecular 
volume  as  does  its  weight  to  the  gram-molecular  weight. 
Thus  the  ratio  2.02/3.09,  between  the  known  weights,  will 
be  exactly  equal  to  the  ratio  between  the  corresponding 
volumes,  —  2.02/3.09  =  22,400/z.  By  simple  proportion 
we  should  have  2.02: 3.09  =  22,400:  x;  or,  with  reference  to 
the  fractional  part  of  the  volume  occupied  by  2.02  grams 
of  the  gas,  we  know  that  3.09  grams  must  occupy  3.09/2.02 

times  this  known  volume,  i.e.,  3>Q9  X  22?4QO  Or  34  261  c.c. 

The  value  of  x,  as  here  calculated,  must  be  adjusted  now 
to  the  observed  conditions  of  temperature  and  pressure. 
For  this  we  need  but  a  moment's  glance  to  see  that  the 
corrected  volume,  x',  will  be  expressed  by  the  equation 


88  CHEMICAL   CALCULATIONS 

The  reverse  process  —  that  of  determining  the  amount 
of  any  substance  which  will  give  a  definite  volume  of  gas 
measured  at  certain  definite  conditions  —  is  made  clear 
when  the  weight  of  this  gas  is  ascertained  (cf.  Ex.  17). 
So  also  the  consideration  of  aqueous  vapor,  when  present, 
brings  into  these  calculations  only  those  methods  already 
outlined  in  previous  examples. 

Calculations  with  Reference  to  Degree  of  Purity.  — 
Oftentimes  the  purity  of  materials  employed  in  chemical 
operations  does  not  come  up  to  the  standard  or  100  per 
cent.  In  such  cases,  where  the  deviation  from  the  theo- 
retical is  known,  the  results  found  must  be  recalculated 
with  reference  to  the  standard  purity.  For  instance,  if 
the  magnesium  considered  in  Example  37  had  been  of 
only  80  per  cent  purity  (contaminated,  we  shall  say,  with 
20  per  cent  of  inert  or  extraneous  matter),  then  our 
results,  wrongly  based  upon  a  valuation  of  100  instead  of 
the  actual  80,  must  be  raised  in  accordance  with  the  ratio 
existing  between  the  actual  and  theoretical  value,  i.e., 
according  to  the  ratio  80/100.  The  result  in  the  exam- 
ple cited  would  be  changed  through  the  simple  pro- 
portion: 120.4  :  x  =  80  :  100;  or  150.5  grams  would  be  the 
weight  of  magnesium  (80  per  cent)  necessary  to  give 
10  grams  of  hydrogen. 

Conversely,  if  it  were  desired  to  ascertain  what  weight 
of  hydrogen  could  be  obtained  from  this  weight  (150.5 
grams)  of  magnesium  (80  per  cent  pure),  then  the  result 
upon  the  basis  of  150.5  grams  of  pure  magnesium  would 
need  to  be  lowered  in  the  same  proportion:  100  :  80. 
The  same  end  is  attained  by  bringing  into  the  original 
calculation  just  80  per  cent  of  the  weight  of  magnesium 
(80  per  cent  pure),  i.e.,  120.4  grams  in  this  example. 

These  calculations  afford,  therefore,  a  simple  means  for 
estimating  the  degree  of  purity  of  a  definite  weight  of  a 


CHEMICAL   EQUATIONS  89 

substance  when  the  amount  of  some  constituent  con- 
cerned in  one  of  its  reactions  is  referred  to  the  correspond- 
ing theoretical  value. 

Example  40'  —  A  specimen  of  marble  weighing  5  grams 
evolved  2.1  grams  of  carbon  dioxide  (corresponding  to  a  theo- 
retical volume  of  1162.5  c.c.  at  20°  and  750  mm.)  when  acted 
upon  by  an  acid.  Estimate  the  degree  of  purity,  assuming  that 
all  of  the  carbonate  was  present  as  calcium  carbonate. 

The  reaction  with  hydrochloric  acid  is  here  given: 
CaCO3  +  2  HC1  =  CaCl2  +  H2O  +  CO2. 
100.09  44 

From  this  equation  the  proportionality  between  the 
reaction-quantities  CaC03  (100.09)  and  CO2  (44)  leads  to 

100  09       5 
the  equation  — ^—  =  —  >  where  x,  2.2  grams,  is  the  weight 

44  X 

of  carbon  dioxide  theoretically  possible  from  this  weight 
of  pure  calcium  carbonate.  The  specimen  is  accordingly 

21 

oaly  —  ,  or  95.45  per  cent,  pure  (2.2  :  2.1  =  100  :  x).    Calcu- 
zz 

lating  from  the  standpoint  of  the  carbon  dioxide,  we  natu- 

44          2  1 

rally  reach  the  same   value :  =  —  ;  or  x,    4.772 

lou.uy      x 

grams,  is  the  theoretical  amount  of  calcium  carbonate  re- 
quired for  the  production  of  2.1  grams  of  carbon  dioxide. 

4772 

This  weight  is  ,  or  95.45  per  cent,  of  the  weight  of  the 

oUUO 

specimen;  consequently,  just  this  fraction  of  the  specimen 
can  be  considered  calcium  carbonate,  i.e.,  its  degree  of 
purity  is  95.45  per  cent. 

Calculation  of  Products  Resulting  from  Mixtures. — 
The  definite  weights  of  various  substances  brought 
together  for  chemical  combination  are  rarely  ever  present 
in  the  exact  proportions  indicated  by  the  equation 
representing  the  particular  reaction. 


90  CHEMICAL   CALCULATIONS 

Thus,  in  the  interaction  between  a  metal  and  an  acid, 
as  already  noted,  the  acid  is  taken  in  excess  of  the  theo- 
retical amount  proportional  to  the  definite  weight  of 
metal  concerned.  Such  an  excess  is  easily  removed 
from  the  final  product  by  volatilization.  In  cases  where 
this  cannot  be  accomplished  without  decomposition  of 
some  product  sought,  other  means  (crystallization,  etc.) 
are  employed.  In  order  to  ascertain  the  exact  amount 
of  any  substance  formed  through  one  of  these  inter- 
actions it  is  necessary  first  to  determine  what  particular 
component  or  components  can  be  acted  upon  to  com- 
pletion by  the  others  present.  This,  of  course,  presupposes 
the  tendency  for  the  reaction  to  run  to  completeness  in 
some  one  direction,  and  implies  the  elimination  of  those 
products  which  may  lead  to  a  reversal  of  these  conditions. 

Example  4!  —  What  weight  of  sodium  sulphate  may  be 
expected  to  result  from  the  interaction  of  10  grams  of  sodium, 
11  grams  of  sulphur  and  40  grams  of  oxygen? 

The  equation  representing  the  possible  combination  of 

these  elements  to  this  end  is  as  follows : 

2  Na  +  S         +   2  02  =  Na2S04 
46       +  32.07  +    64    =  142.07 

Weights  given.  {10  1 1  40. 

The  actual  weights  of  the  several  substances  here 
entering  into  combination  as  sodium  sulphate  must 
stand  in  the  same  relation  to  each  other  as  do  the  cor- 
responding reaction-quantities.  There  is  then  an  equiva- 
lence in  the  ratios  between  each  reaction-quantity  and  the 
corresponding  weight  that  represents  it  in  the  reaction 
(see  footnote,  page  72). 

If  these  conditions  are  fulfilled  in  the  example,  we 
shall  have  an  equality  in  the  ratios,  46/10,  32.07/11 
and  64/40.  A  single  glance,  however,  disproves  this 


CHEMICAL  EQUATIONS  91 

point.  It  then  becomes  necessary  to  determine  by  trial 
which  one  of  these  ratios,  between  a  reaction-quantity 
and  its  corresponding  weight  as  given  in  the  example,  is 
the  correct  ratio  to  select  as  the  basis  of  calculation.  For 
example,  upon  the  basis  of  64/40,  we  must  have: 
64  =  46  =  32.07  where  (2g7)  and  (2Q)  ag  the  re_ 
40  z  y 

spective  values  for  sodium  and  sulphur,  actually  exceed 
the  amounts  at  hand.  Upon  the  basis  32.07/11,  we  have: 

§M?  =  i?  =  M,  where  x,  (15.7),  and   y,  (22),  are   the 
11          x        y 

respective  values  for  sodium  and  oxygen.  In  this  latter 
case  only  the  sodium  is  higher  than  the  weight  stipu- 
lated. Finally,  upon  the  basis  46/10,  we  have  the  equa- 
tion —  =  ?^Z  =  ?4 ,  and  from  this  both  x,  (6.97),  the 
x  y 

weight  of  sulphur,  and  y,  (13.9),  the  weight  of  oxygen, 
fall  below  the  amounts  given;  hence  46/10  is  the  proper 
ratio  upon  which  to  base  the  calculation.  The  sum  of 
the  respective  weights  thus  obtained,  30.87  grams,  (10  + 
6.97  +  13.97),  gives  the  highest  weight  of  sodium  sulphate 
possible  from  the  data  in  the  problem. 

In  general  we  select  as  the  basis  of  calculation  for  any 
given  reaction  only  that  ratio  which  brings  into  consid- 
eration those  values  for  the  various  substances  involved 
as  do  not  exceed  the  amounts  present.  More  simply 
stated,  perhaps,  we  base  our  calculations  upon  that  weight 
of  a  particular  substance  at  hand  which  gives,  with  its 
corresponding  reaction-quantity  as  denominator,  the 
smallest  numerical  factor.  Thus,  in  the  example  above: 
10  .  (11)  . (40)  x 

46  :  32.07  :  64   ==  142.07* 

Under  conditions  where  reversible  reactions  are  likely, 
for  example  the  preparation  of  sodium  sulphate  by  the 


92  CHEMICAL   CALCULATIONS 

action  of  sulphuric  acid  upon  common  salt,  as  shown  in 
the  equation: 

2  Nad  +  H2S04  =  Na2SO4  +  2  HC1, 
we  determine  in  similar  manner  the  weight  of  sodium 
sulphate  possible  from  any  known  weight  of  salt  (if  the  sul- 
phuric acid  is  in  excess)  or  from  the  weight  of  sulphuric 
acid  (if  the  salt  is  in  excess).  The  presence  of  hydro- 
chloric acid  as  the  reversing  agent  must  be  removed  in 
either  case  if  we  wish  to  obtain  the  calculated  results. 

Calculation  of  Ratios  between  Reaction-Quantities  in 
Dependent  Equations.  —  When  an  element  or  group  of 
elements  enters  into  a  series  of  successive  reactions  and 
the  equation  for  each  reaction  can  be  constructed,  we 
may  draw  up  a  proportionality  between  the  reaction- 
quantities  in  any  two  of  the  equations  providing  that 
some  quantity  is  common  to  both.  In  like  manner  this 
proportionality  may  be  extended  step  by  step  over  any 
number  of  dependent  equations  (cf.  Ex.  24). 

Example  4&-  —  What  weight  of  bromine  can  be  liberated 
from  a  concentrated  solution  of  potassium  bromide  (excess) 
by  the  addition  of  12  grams  of  hydrochloric  acid  (containing 
39.1  per  cent  HC1)  and  an  excess  of  manganese  dioxide? 

The  two  equations  representing  the  action  are  as  fol- 
lows: 

(a)   Mn02   +  4  HC1  =  MnCl2  +  C12  +  2  H2O 
(6)   2  KBr  +  C12       =2  KC1  +  Br2. 

It  is  observed  that  the  reaction-quantity  C12  is  common 
to  both  (a)  and  (6),  consequently  the  reaction-quantities 
directly  proportional  to  this  quantity  in  either  equation 
will  be  also  directly  proportional  to  each  other.  The 
ratio  4  HC1/C12  from  equation  (a)  and  the  ratio  Cl2/Br2 
from  equation  (6)  give  us,  accordingly,  the  ratio  4  HCl/Br3 
for  the  proportionality  between  the  reaction-quantities 


CHEMICAL  EQUATIONS  93 

required  for  this  example.  From  the  equivalence  between 
this  ratio  and  the  ratio  of  the  weights  corresponding 
thereto,  we  have  4  HCl/Br2,  or  4(36.47) /2(79.92),  or 
145.88/159.84  =  12 /x,  which,  solved  for  x,  gives  a  value 
of  13.1  grams.  This  weight  of  bromine,  13.1  grams,  is 
based  upon  the  hydrochloric  acid  as  100  per  cent  hydro- 
gen chloride.  The  concentrated  acid  at  our  command, 
the  acid  stated  in  the  example,  contained  only  39.1  per 
cent  hydrogen  chloride.  It  remains  then  to  calculate  the 
weight  of  bromine  which  a  39.1  per  cent  acid  can  give. 
By  reference  to  Example  40,  the  method  is  outlined  to 
be  simply  one  of  proportion,  according  to  which  we  shall 
have  100  :  39.1  =  13.1  :  x, 

or  z  =  5.13  grams,  the  weight  of  bromine  evolved  by  12 
grams  of  39.1  per  cent  hydrochloric  acid.  This  same 
result  is  easily  obtained  by  determining  first  the  weight 
of  hydrogen  chloride  in  the  12  grams  of  39.1  per  cent  acid, 
(12  X  39.1  per  cent  =  4.69),  and  working  the  example 
with  this  value  for  the  hydrogen  chloride. 

The  elimination  of  a  quantity  occurring  in  two  depend- 
ent equations  may  necessitate  a  readjustment  of  one  or 
both  of  the  equations  containing  it  before  the  quantity 
becomes  alike  in  each. 

Example  4s-  —  What  weight  of  iron  sulphide  will  be  required 
to  furnish  sufficient  hydrogen  sulphide  for  the  reduction  of  10 
grams  of  sulphur  dioxide  to  sulphur? 

The  two  equations  here  required  are  as  follows  : 
(a)   FeS     +  2  HC1  =  FeCl2   +  H2S 
(6)    2  H2S  +  SO2     =2  H2O  +  3  S. 

The  reaction  in  equation  (b)  depends  upon  the  hydro- 
gen sulphide  that  is  evolved  in  (a);  consequently  this  sub- 
stance must  constitute  the  common  reaction-quantity. 
In  order  to  make  this  quantity  alike  for  the  two  equations 


94  CHEMICAL   CALCULATIONS 

and  thus  eliminate  it  from  the  calculations,  it  is  only  neces- 
sary to  multiply  equation  (a)  by  2,  when  we  obtain  (a') : 
(a')     2  FeS  +  4  HC1  =  2  FeCl2  +  2  H2S. 

The  comparison  of  any  of  the  reaction-quantities  in  the 
two  equations  (a')  and  (6)  with  reference  to  hydrogen  sul- 
phide is  now  made  simple.  In  the  example  given,  iron 
sulphide  and  sulphur  dioxide  are  found  to  be  related 
through  the  ratios  2  FeS/2  H2S  and  2  H2S/S02,  or  directly 
as  2  FeS/S02.  The  calculation  is  conducted,  therefore, 
as  follows : 
2  FeS/S02,  or  2  (87.92) /64.07,  or  175.84/64.07  =  x/ 10. 

From  which  x,  the  weight  of  iron  sulphide,  is  found  to  be 
27.44  grams. 

Without  this  elimination  of  the  quantity  common  to  both 
equations,  the  calculation,  of  course,  can  be  made  directly 
toward  ascertaining  the  weight  representing  this  quan- 
tity in  the  first  equation,  and  then  finally,  from  this  weight, 
the  weight  representing  any  other  reaction-quantity  pro- 
portional to  it  in  the  second  equation.  Such  calculations, 
here  involving  the  weight  of  hydrogen  sulphide,  are  indeed 
roundabout  and  entirely  unnecessary. 

Calculation  of  Ratios  between  Reaction-Quantities  in 
Independent  Equations.  —  In  the  study  of  dependent 
equations  the  reaction-quantities  may  be  regarded  as 
related  to  each  other,  through  this  common  reaction-quan- 
tity, as  are  the  members  of  a  single  equation.  The  reac- 
tion-quantities of  the  second  equation  are,  so  to  speak, 
brought  into  existence  through  the  agency  of  this  com- 
mon quantity.  In  the  study  of  independent  equations, 
where  there  is  present  no  one  quantity  which  has  a  direct 
bearing  upon  any  other  equation,  we  have  simply  a  fur- 
ther application  of  this  same  principle;  namely,  the  com- 
parison of  all  the  possible  reaction-quantities  in  the 


CHEMICAL  EQUATIONS  95 

separate  equations  so  long  as  some  quantity  can  be  made 
common  to  each.  Quantities  so  compared  will  be  directly 
proportional  to  each  other,  but  only  in  respect  to  this 
common  reaction-quantity.  As  an  illustration,  the  fol- 
lowing equations  are  cited: 

(a)   Na2C03    +  2  HC1  =  2  Nad  +  H2O  +  C02 
(6)    NaHC03  +  HC1     =  Nad     +  H2O  ±  CO2. 

In  these  independent  equations  there  are  a  number  of 
substances  represented  for  which  the  reaction-quantities 
could  be  adjusted  alike  for  both;  thus  the  reaction-quan- 
tity CO2  is  a  common  one.  Upon  this  fact  we  may  draw 
up  the  ratio  Na2C03/NaHC03  to  express  the  proportion- 
ality between  the  relative  amounts  of  normal  carbonate  and 
primary  carbonate  necessary  to  give  an  equal  amount  of 
carbon  dioxide,  i.e.,  the  relation  is  based  upon  the  carbon- 
dioxide  content.  In  order  to  obtain  a  comparison  with 
reference  to  the  salt,  equation  (6)  must  be  doubled  to 
(60: 

(&0   2  NaHC03  +  2  HC1  =  2  NaCl  +  2  H2O  +  2  C02. 

The  ratio  Na2C03/2  NaHCO3  then  expresses  the  relation 
between  the  relative  amounts  of  each  carbonate  neces- 
sary to  give  equal  amounts  of  salt  with  hydrochloric  acid. 
Or,  since  sodium  is  always  a  constant  quantity  in  salt, 
we  may  say  that  the  ratio  above  is  that  based  upon  a  like 
content  of  sodium  in  each  carbonate. 

In  addition  to  the  carbonate  discussed  in  the  preceding 
paragraph,  we  may  also  compare  the  reaction-quantities 
for  the  carbon  dioxide  present.  Between  equations  (a) 
and  (6),  where  this  is  a  common  quantity,  we  have  the 
ratio  2  HC1/HC1  representing  the  relative  amounts  of  acid 
necessary  in  the  respective  cases  to  give  equal  amounts 
of  carbon  dioxide.  Between  equations  (a)  and  (6X)  we 
have  the  ratio  CO2/2  CO2  representing  the  relative 


96  CHEMICAL   CALCULATIONS 

amounts  of  carbon  dioxide  evolved  from  equal  amounts 
of  sodium  when  contained  respectively  in  normal  car- 
bonate or  primary  carbonate,  as  indicated  by  the  ratio 
Na2CO3/2  NaHCOg,  or  the  ratio  2  NaCl/2  NaCl,  in  each  of 
which  the  sodium  content  is  the  same  for  both  terms  of 
the  ratio.  This  latter  point  may  be  illustrated  by  the 
equation: 

2  NaHCO3  =  Na2CO3  +  CO2  +  H2O. 

Here  it  is  shown  that  two  molecules  of  the  primary  car- 
bonate break  down  into  one  molecule  of  the  normal  car- 
bonate with  the  loss  of  one  molecule  of  carbon  dioxide. 
The  ratio  2  NaHCO3/CO2  represents  the  relative  amounts 
of  these  substances  concerned  in  the  action  of  heat  upon 
the  primary  carbonate.  This,  in  fact,  is  a  case  where  the 
entire  quantity  of  a  substance,  (CO2),  available  in  a  com- 
pound need  not  be  concerned  in  a  ratio  for  the  study  of 
that  compound,  but  only  that  portion  of  it  separately  in- 
volved as  a  reaction-quantity  and  made  directly  propor- 
tional, therefore,  to  some  other  reaction-quantity. 

Example  44-  —  What  relative  weights  of  mercuric  oxide  and 
barium  peroxide  are  required  in  the  preparation  of  equal 
amounts  of  oxygen? 

The  molecular  equations  with  oxygen  as  the  common 
quantity  adjusted  alike  in  both  are  as  follows: 

2  HgO  =  2    Hg   +  O2 
2  BaO2  =  2  BaO  +  O2. 

The  ratio  2  HgO/2  Ba02,  or  HgO/BaO2,  or  216/169.37, 

determines  accordingly  the  relation  between  the  corres- 
ponding weights  of  these  two  substances  in  this  problem. 
If  we  take  one,  e.g.,  the  mercuric  oxide,  as  100  we  re- 
duce the  second  to  a  comparatively  simple  value: 

216      =  100 
169.37  ~     x 


CHEMICAL   EQUATIONS  97 

This  gives  78.4  as  the  value  of  x,  or  that  weight  in 
grams  of  barium  peroxide  equal  to  100  grams  of  mercuric 
oxide  in  the  preparation  of  oxygen. 

Calculation  of  Reaction-Quantities  from  the  Weights 
of  Substances  Involved.  —  The  direct  proportionality 
which  exists  between  the  reaction-quantities  of  a  given 
equation  requires,  as  we  have  seen,  the  same  proportion- 
ality between  the  corresponding  weights  which  represent 
them  in  this  particular  reaction.  This  permits  of  an 
equivalence  in  the  ratios  between  each  reaction-quantity 
and  its  corresponding  weight  (cf.  Ex.  41  and  footnote, 
page  72). 

In  the  construction  of  chemical  equations  from  the 
actual  weights  of  substances  therein  concerned,  we  must 
bear  in  mind  the  possibility  of  deviations  in  these  weights 
from  those  demanded  in  the  reactions.  These  deviations 
may  be  due  to  any  number  of  causes  and  rapidly  increase 
with  the  instability  of  the  compounds  considered  as  well 
as  with  the  tendencies  for  secondary  reactions.  Conse- 
quently the  determination  of  the  correct  reaction-quan- 
tities for  these  equations  must  be  made  a  special  study  in 
each  individual  case. 

Example  45.  —  A  solution  of  10  grams  of  crystallized  sodium 
thiosulphate,  Na2S203 . 5  H2O  decolorized  5.1  grams  of  iodine. 
What  molecular  quantity  of  this  salt  was  associated  with  one 
molecule  of  iodine  in  this  reaction? 

From  the  equivalence  in  ratios  between  the  reaction- 
quantities  and  the  actual  weights  involved  we  have: 

£_      _!,  x_      253.84 

10      5.1 '  Cr  10*     5.1 

This  gives  to  x  the  value  497.7,  the  reaction-quantity  of 
the  thiosulphate  associated  with  one  molecular  weight  of 


98  CHEMICAL   CALCULATIONS 

iodine.  The  molecular  weight  of  sodium  thiosulphate, 
Na2S203 . 5  H20,  is  248.2,  consequently  we  have  here 
497.7/248.2  or  approximately  2  molecules  of  this  salt  in 
its  reaction-quantity,  i.e.,  the  equation  constructed  upon 
the  iodine  involved  as  just  one  molecule  will  be 

2  Na2S203 .  5  H20  +  I2  =  (2  Nal  +  Na2S406). 
Complex  Reaction-Quantities.  —  There  are  a  number  of 
equations  in  which  the  reaction-quantities  upon  one  side 
are  incorporated  into  one  single  reaction-quantity  upon 
the  other  side.  The  same  principles  hold  here  as  in  the 
cases  just  discussed,  but  the  study  of  this  larger  quantity 
is  nothing  more  or  less  than  the  study  of  the  formula- 
quantities  present  in  it.  The  consideration  of  the  so-called 
molecular  compounds,  as  are  the  double  salts  and  salts 
containing  water  of  crystallization,  illustrates  this  point. 

Example  46.  —  100  grams  of  copper  sulphate  will  give  what 
weight  of  blue  vitriol  (CuS04 .  5  H2O)? 
From  the  equation,  — 

CuSO4  +  5  H20  =  CuSO4 .  5  H2O 
159.64  +  90.1      =  249.74, 
the  ratio 

CuSO4  159.64 

CuSO,.5H20'  Cr  249.74 

is  a  constant,  and  denotes  the  direct  proportionality 
between  these  two  reaction-quantities.  The  ratio  between 
the  actual  weights,  100  and  x,  when  put  equal  to  the  ratio 
between  the  molecular  quantities,  gives  us 

159.64  =  100 

249.74        x   ' 

where  x,  with  the  value  156.4  grams,  is  the  weight  of  blue 
vitriol.  As  the  percentage  composition  of  the  anhydrous 
copper  sulphate  is  a  constant,  we  may  just  as  well  calcu- 
late what  weight  of  pure  copper,  Cu,  or  sulphur,  S,  or  even 


CHEMICAL   EQUATIONS  99 

oxygen,  4  O,  is  necessary  to  give  any  definite  weight  of 
anhydrous  copper  sulphate  and  finally  blue  vitriol;  or, 
vice  versa,  what  weight  of  blue  vitriol  is  obtainable  from 
any  definite  weight  of  one  of  these  constituents. 

As  another  illustration  of  this  point  we  may  take  the 
formation  of  ordinary  alum: 


A12(SO4)3 .  18  H2O  =  KjSO,  .  Ala(SO4)3 .  24H,O 
1  mol.  +  1  mol.  =          1  mol. 

A  molecular  quantity  of  one  sulphate  unites  with  one  of 
another  to  form  one  molecular  quantity  of  the  double 
sulphate.  These  quantities  are  all  directly  proportional 
to  each  other,  and  from  the  equation  we  write  the  following 
ratios: 

(a\  K2S04.6H20 

K2SO4 .  A12(SO4)3 .  24  H2O 

(6x  A12(SO4)3 .  18  H2O 

V  ;     K2SO4 .  A12(SO4)3 .  24  H2O 

,  v  K2S04 .  6  H2O 

A12(SO4)3.18H2O* 

These  serve  as  the  ratios  for  calculating,  upon  the  molecu- 
lar quantities  involved,  the  amount  of  alum  obtainable 
from  certain  known  amounts  of  (a)  crystallized  potassium 
sulphate;  (b)  crystallized  aluminium  sulphate;  and  also 
(c)  the  amount  of  crystallized  aluminium  sulphate  neces- 
sary for  combination  with  one  molecule  of  the  crystallized 
potassium  sulphate,  or  vice  versa.  Though  somewhat 
more  complicated  molecular  aggregates  may  be  present, 
the  ratios  between  the  several  factors  are  always  constant. 

Example  47.  —  How  much  alum  can  be  prepared  from  100 
grams  of  anhydrous  potassium  sulphate? 

The  ratio 

K2S04 174.27 

K2SO4 .  A12(SO4)3 .  24  H2O  '  °r  949.16 


100  CHEMICAL   CALCULATIONS 

is  placed  equal  to  the  ratio  100/£,  and  the  equation  solved 
in  the  usual  manner.  The  value  of  x,  the  weight  of  alum, 
is  found  to  be  544.6  grams. 


13, 


PROBLEMS. 


What  weight  of  potassium  hydroxide  may  be  prepared 
by  the  action  of  100  grams  of  potassium  upon  water? 

Am,  143.5  grams. 

138.  What  weight  of  potassium  will  be  required  in  the  prep- 
aration of  20  grams  of  potassium  carbonate,  K2C03? 

Ans.  11.3  grams. 

139.  What  weight  of  magnesium  chloride,  MgCl2,  may  be 
obtained  by  the  action  of  hydrochloric  acid  upon  10  grams  of 
magnesium  carbonate,  MgC03?     What  weight  of  carbon  dioxide 
will  be  liberated?  Ans.  11.3  grams  MgCla. 

5.2  grams  C0a. 

140.  Recalculate  Problem  139  on  the  supposition  that  the 
magnesium  carbonate  contained  10  per  cent  of  insoluble  matter. 

Ans.  10.16  grams  MgCla. 
4.7  grams  C0a. 

141.  What  weight  of  sulphur  dioxide,  SO2,  can  be  obtained 
by  the  action  of  an  acid  upon  250  grams  of  sodium  sulphite, 
Na2S03?  .  Ans.  127  grams. 

142.  Recalculate  Problem  141  on  the  supposition  that  20 
per  cent  of  the  sulphite  had  become  oxidized  to  sulphate. 

Ans.  101.6  grams. 

143.  Calculate  the  volume  of  carbon  dioxide,  at  22°  and  740 
mm.  pressure,  that  will  be    liberated  by  the  action  of   acid 
upon  20  grams  of  calcium  carbonate,  CaCO3.       Ans.  4967  c.c. 

^  144.  What  weight  of  magnesium  will  be  required  for  the 
liberation  of  500  c.c.  of  hydrogen,  at  20°  and  740  mm.  pressure, 
when  acted  upon  by  an  acid?  Ans.  0.49  gram. 

V/  145.  Calculate  the  volume  of  hydrogen,  measured  over  water 
at  17°  and  742  mm.  pressure,  that  can  be  liberated  by  the 
action  of  10  grams  of  sodium  upon  water.  Ans.  5403  c.c. 

146.   What  weight  of  aluminium  will  be  required  for  the 
liberation  of  420  c.c.  of  hydrogen,  measured  over  water  at  18° 


CHEMICAL  £^tf&#foKs}  ;J/  >\\  \  ;VJ,    101 


and  746.4  mm.  pressure,  when  acted  upon  by  hydrochloric 
acid?  Ans.  0.306  gram. 

147.  What    weight    of   ammonium    nitrite,    NH4N02,    will 
evolve,  when  heated,  480  c.c.  of  nitrogen,  measured  over  water 
at  21°  and  747.5  mm.  pressure?  Ans.  1.22  grams. 

148.  Determine  the  purity  of  a  sample  of  anhydrous  sodium 
carbonate,  Na2CO3,  6  grams  of  which  gave,  when  acted  upon  by 
an  acid,  1310  c.c.  of  carbon  dioxide,  at  10°  and  750  mm.  pressure. 

Ans.  98.35  per  cent  pure. 

149.  Determine  the  purity  of  a  sample  of  anhydrous  sodium 
carbonate,  3  grams  of  which  gave,  by  treatment  with  sulphuric 
acid  and  final  ignition,  3.99  grams  of  sodium  sulphate,  Na2S04. 

Ans.  99.25  per  cent  pure. 

150.  A  sample  of  iron  wire  weighing  2.4  grams  was  found  to 
give,  when  acted  upon  by  excess  of  acid,  a  volume  of  hydrogen 
measuring  1015.1  c.c.,  at  10°  and  745  mm.  pressure.     What  was 
its  degree  of  purity?    The  presence  of  any  other  substance 
capable  of  liberating  hydrogen  from  an  acid  is  here  disregarded. 

Ans.  99.72  per  cent  pure. 

151.  A  sample  of  silver  nitrate  weighing  2.40  grams  was 
brought   into    solution   and   treated   with   a    soluble    chloride 
(excess).     The  weight  of  silver  chloride,  AgCl,  precipitated  was 
2.01  grams.     What  was  the  purity  of  the  sample? 

Ans.  99.26  per  cent  pure. 

152.  What  weight  of  zinc  (98  per  cent  pure)  will  be  required 
for  the  liberation  of  the  hydrogen  from  10  grams  of  hydro- 
chloric acid  containing  39.1  per  cent  HC1?    Ans.  3.576  grams. 

153.  What  weight  of  sulphuric  acid  containing  27.32  per 
cent  H2SO4  will  be  required  for  interaction  with  2.17  grams  of 
iron  wire  (99  per  cent  pure)  ?  Ans.  13.82  grams. 

154.  What  volume  of  hydrogen,  measured  over  water  at  18° 
and  746.4  mm.  pressure,  will  be  liberated  by  the  action  of 
aluminium  upon  20  grams  of  sulphuric  acid  containing  41.  £ 
per  cent  H2S04?  Ans.  2100  c  c. 

155.  What  weight  of  sulphuric  acid  (27.32  per  cent  H2S04) 
will  be  required  for  interaction  with  a  metal  (Zn,  Mg,  etc.)  in 
order  to  give  a  volume  of  hydrogen  measuring  over  water  103* 
c.c.,  at  16°  and  742.5  mm.  pressure?  Ans.  15  grams. 


102 


CALCULATIONS 


156.  What  weight  of  hydrochloric  acid  (23.82  per  cent  HC1) 
will  be  required  for  interaction  with  iron  sulphide,  FeS,  in  order 
to  give  a  volume  of  hydrogen  sulphide,  measuring  613.1  c.c.,  at 
20°  and  745  mm.  pressure? 

Calculate  also  the  weight  of  iron  sulphide  (98  per  cent  pure) 
here  consumed.  Ans.  7.655  grams  acid. 

2.243  grams  FeS  (98  per  cent). 

157.  When  100  grams  of  mercury  and  20  grams  of  sulphur 
are  rubbed  together  what  weight  of  mercuric  sulphide,  HgS, 
may  be  formed?  Ans.  116  grams. 

Note.  —  From  the  reaction-quantities  involved  it  is  seen  that  the 
sulphur  is  in  excess.  This  excess  is  easily  removed  by  solution  in 
carbon  disulphide. 

158.  A  mixture  of  10  grams  of  zinc  dust  and  2  grams  of 
sulphur  was  gently  heated  to  point  of  reaction.     What  weight 
of  zinc  sulphide,  ZnS,  was  formed?  Ans.  6.053  grams. 

159.  A  mixture  of  10  grams  of  iron  and  8  grams  of  sulphur 
was  gently  heated  to  point  of  reaction.      What  weight  and 
volume  (at  standard  conditions)  of  hydrogen  sulphide  could  be 
obtained  from  the  final  product,  FeS,  by  the  action  of  an  acid? 

Ans.  6.104  grams,  or  4010.5  c.c.  H2S. 

160.  A  mixture  of  4  grams  of  sodium  oxide,  Na20,  and  6 
grams  of  sulphur  trioxide  will  give  what  weight  of  sodium 
sulphate?  Ans.  9.166  grams. 

161.  3  grams  of  silver  nitrate,  AgN03,  and  1  gram  of  potas- 
sium chloride,  KC1,  were  brought  together  in  aqueous  solution. 
What  weight  of  silver  chloride,  AgCl,  was  precipitated? 

Ans.  1.923  grams. 

162.  8.2  grams  of  crystallized  barium  chloride,  BaCl2 .  2H20, 
and  7  grams  of  sulphuric  acid  (70  per  cent  H2S04)  were  brought 
together  in  aqueous  solution.     What  weight  of  barium  sulphate, 
BaS04,  was  precipitated?  Ans.  7.83  grams. 

163.  10  grams  of  a  mixture  of  marble  (CaC03),  magnesium, 
an(J  an  inert  substance  were  acted  upon  by  excess  of  acid.     The 
carbon  dioxide  evolved  (taken  up  in  a  solution  of  potassium 
hydroxide)  was  found  to  weigh  1.318  grams.    The  volume   of 
the  other   gas,  hydrogen,  measured  over  water  at    10°    and 


CHEMICAL   EQUATIONS  103 

750.2  mm.  pressure,  was  5874  c.c.     Calculate  the  percentage 
amount  of  each  component  in  the  mixture. 

Ans.  30  per  cent  marble. 

60  per  cent  magnesium. 

10  per  cent  insoluble  matter. 

164.  15  grams  of  an  alloy  of  zinc  and  copper  (containing 
10  per  cent  of  copper)  were  placed  in  a  vessel  containing  100 
grams  of  sulphuric  acid  (25  per  cent  H2SO4).     What  weight  of 
hydrogen  was  liberated?  Ans.  0.417  gram. 

165.  12  grams  of  an  alloy  of  aluminium  and  zinc  (containing 
33J  per  cent  of  zinc)  were  placed  in  a  vessel  containing  180 
grams  of  hydrochloric  acid  (35  per  cent  HC1).     What  volume  of 
hydrogen,  at  standard  conditions,  was  liberated? 

Ans.  11,290  c.c. 

166.  A  specimen  of  silver,  containing  3  per  cent  copper, 
weighed  9.8  grams.     After  solution  in  nitric  acid  an  excess  of 
sodium  chloride  was  added  to  it.     Calculate  the  weight  of  the 
silver  chloride  precipitated.  Ans.  12.632  grams. 

167.  10  grams  of  phosphorus  tribromide,  PBr3,  were  mixed 
with  an  excess  of  water  and  an  excess  of  silver  nitrate  added  to 
this  solution.     What  weight  of  silver  bromide   was   formed? 
Calculate  also  the  exact  weight  of  silver  nitrate  required  for 
this  removal  of  bromide.  Ans.  20.81  grams  AgBr. 

18.82  grams  AgN03. 

168.  What  volume  of  hydrogen  sulphide,  at  standard  con- 
ditions, would  be  required  for  interaction  with  an  excess  of 
iodine,  in  aqueous  suspension,  in  order  to  furnish  an  amount 
of  hydriodic  acid  sufficient  for  the  precipitation  of  10  grams  of 
silver  iodide  from  a  solution  of  silver  nitrate?     Ans.  477  c.c. 

169.  What  weight  of  fluorspar,  CaF2,  would  be  required,  to 
furnish  sufficient  hydrogen  fluoride   (by  interaction  with   sul- 
phuric acid)  to  convert  5  grams  of  quartz,  SiO2,  into  silicon 
fluoride,  SiF4?  Ans.  12.95  grams. 

170.  Calculate  the  volume  of  chlorine,  at  standard  condi- 
tions, necessary  to  give,  by  interaction  with  water,  an  amount 
of  oxygen  which  will  just  suffice  for  the  oxidation  of  10  grams 
of  mercury  to  mercuric  oxide,  HgO.  Ans.  1120  c.c. 


104  CHEMICAL   CALCULATIONS 

171.  Calculate  the  volume  of  chlorine,  at  standard  condi- 
tions, necessary  to  give  an  amount  of  potassium  chlorate   (by 
interaction  with  a  hot  solution  of  potassium  hydroxide)  which 
would  just  suffice,  in  its  decomposition  into  oxygen  and  potas- 
sium chloride,  for  the  oxidation  of  5  grams  of  hydrogen  to 
water.  Ans.  55,446  c.c. 

172.  100  grams  of  iron  pyrites,  FeS2,  were  roasted  to  ferric 
oxide,  Fe2O3,  and  sulphur  dioxide.     The  sulphur  dioxide  was 
then  taken  up  by  sodium   peroxide,  Na2O2,  to  form  sodium 
sulphate,  and  this  product  treated  with  a  solution  of  barium 
chloride,  BaCl2.     What  weight  of  barium  sulphate  was  pre- 
cipitated? Ans.  389.1  grams. 

173.  20  grams  of  nitrogen  were  carried  through  the  follow- 
ing  series    of    reactions.     Calculate   the    resulting   volume    of 
nitrous  oxide,  N20,  at  standard  conditions. 

3Mg  +  N2  =  Mg3N2 
Mg3N2  +  6  H2O  -  3  Mg  (OH),  +  2  NH3 
NH3+HNO3  =  NH4NO3 

NH4NO3  -  N2O  +  2  H2O 

Ans.  32,000  c.c. 

174.  Calculate  the  relative  weights  of  sodium  chlorate  and 
potassium  chlorate  necessary  to  give  the  same  volume  of  oxygen. 

Ans.  100  (NaC103) :  115  (KC1O3). 

175.  Calculate  the  relative  weights  of  potassium  chlorate 
and  perchlorate,  KC104,  necessary  to  give  the  same  volume  of 
oxygen.  Ans.  100  (KC103) :  84.8  (KC104). 

176.  Compare  the  weights  of  aluminium  and  zinc  neces- 
sary for  the  production  of  equal  weights  of  hydrogen  by  inter- 
action with  an  acid.  Ans.  100  (Al) :  361.8  (Zn). 

177.  Compare  the  weight  of  calcium  nitride,  Ca3N2  (in  its 
interaction  with  water),  and  the  weight  of  ammonium  chloride 
(in  its  interaction  with  a  base),  necessary  to  give  the  same 
weight  of  ammonia.  Ans.  100  (Ca3N2) :  72.2  (NH4C1). 

178.  What    relative    weights    of    cupric    oxide,    CuO,    and 
cuprous  oxide,  Cu2O,  are  procurable  from  the  same  weight  of 
copper?  Ans.  100  (CuO) :  89.95  (CuaO). 


CHEMICAL   EQUATIONS  105 

179.  In  the  interaction  of  methane,  CH4,  and  chlorine,  2 
grams  of  the  former  required  17.7  grams  of  the  latter.     Calcu- 
late the  reaction-quantity  of  chlorine  per  molecule  of  methane. 

Ans.  2  C12. 

180.  2  grams  of  hydrogen  sulphide,  H2S,  decolorized  5.8 
grams  of  potassium  dichromate,   K2Cr207   (in  acid  solution). 
Calculate   the   reaction-quantity   of   the   former   required   per 
molecule  of  the  latter.  Ans.  3  H2S. 

181.  2.4  grams  of  ammonia,  NH3,  reduced  17  grams  of  hot 
cupric  oxide,  CuO,  to  copper.     Calculate  the  reaction-quantity 
of  cupric  oxide  required  per  molecule  of  ammonia. 

Ans.  1^  CuO. 

NOTE.  —  In  order  to  avoid  fractional  quantities  we  may  here  mul- 
tiply by  two  and  obtain  3  CuO  per  2  NH3. 

182.  What    weight    of    chrome-alum,    K2SO4 .  Cr2  (SO4)3 . 
24  H2O,  may  be  obtained  from  20  grams  of  crystallized  potas- 
sium  sulphate,    K2SO4 .  6  H2O,    and   an   excess   of   chromium 
sulphate?  Ans.  70.76  grams. 

183.  What    weight    of    ammonium-magnesium    phosphate, 
NH4MgP04 .  6  H20,  could  be  formed  from  a  solution  containing 
50  grams  of  crystallized  magnesium  sulphate,  MgS04 .  7  H20, 
and  an  excess  of  ammonia  and  sodium  phosphate? 

Ans.  49.79  grams. 

184.  What   weight   of   iron-ammonium   alum,    (NH4)2SO4 . 
Fe2(SO4)3 .  24  H2O,  may  be  formed  when  12  grams  of  ammo- 
nium sulphate,  (NH4)2SO4,  and  30  grams  of  ferric  sulphate, 
Fe2(S04)3,    are    brought    together    in    concentrated    aqueous 
solution?  Ans.  72.3  grams. 


CHAPTER  X. 
NORMAL   SOLUTIONS. 

IN  reactions  between  substances  in  solution  no  atten- 
tion thus  far  has  been  given  to  the  actual  amount  of 
substance  contained  in  a  definite  volume  of  solvent. 

In  the  action  between  an  acid,  furnishing  hydrogen- 
ion,  and  a  base,  furnishing  hydroxide-ion,  the  point 
of  neutralization  is  reached  when  equal  quantities  of 
these  two  kinds  of  ions  —  chemically  equivalent  —  are 
present,  and  the  complete  removal  of  both  in  the  form 
of  the  compound  water  as  a  slightly  ionized  substance 
is  effected.  The  detection  of  this  point  in  solution  is 
readily  accomplished  through  the  use  of  an  indicator,  or 
some  substance  which  shows  a  change  in  color  by  the 
merest  trace  of  either  the  one  or  the  other  of  these  two 
ions.  The  operation  of  ascertaining  this  end-point  is 
called  titration.  Naturally  it  may  be  applied  to  the 
determination  of  other  end-points,  or  points  of  comple- 
tion of  definite  chemical  reactions  in  solution,  as  well  as 
to  this  process  of  neutralization. 

The  neutralization  of  hydrochloric  acid  by  sodium  hy- 
droxide is  shown  by  the  following  equation: 
Na'  +  OH'   +  H*     +C1'       =  Na'+Cl'      +   H20 

23  +  17.01       1.01  +35.46  =  23  +  35.46       18^ 
40.01  36.47  58.46  18.02* 

A  solution  which  contains  40.01  parts  by  weight  of 
sodium  hydroxide  will  exactly  neutralize  one  which  con- 
tains 36.47  parts  by  weight  of  hydrogen  chloride.  This 

106 


NOBMAL   SOLUTIONS  107 

follows  from  the  fact  that  in  the  former  there  are  17.01 
parts  of  ionizable  hydroxyl  and  in  the  latter  1.01  parts  of 
ionizable  hydrogen,  —  the  exact  proportions  of  these  two 
substances  necessary  for  the  formation  of  water.  When 
equal  volumes  of  these  solutions  neutralize  each  other, 
then  the  concentration  of  ionizable  hydrogen  in  the  acid 
solution  is  equal  to  the  concentration  of  ionizable 
hydroxyl  in  the  solution  of  the  base;  i.e.,  the  relative 
amounts  of  each  are  directly  proportional  to  1.01  and 
17.01  respectively.  The  liter  has  been  adopted  as  the 
standard  volume  for  reactions  in  solution;  when  a  gram- 
molecular  weight  of  a  substance  is  contained  in  this  vol- 
ume we  have  what  is  called  a  gram-molecular  solution  or 
more  commonly  a  Molar  Solution.  Some  definite  tem- 
perature, circa  20°,  is  usually  undei^tood. 

From  the  reaction  between  sodium  hydroxide  and 
hydrochloric  acid,  and  from  the  definitions  just  given,  we 
are  aware  that  one  liter  of  a  molar  solution  of  the  former 
will  exactly  neutralize  one  liter  of  a  molar  solution  of 
the  latter;  consequently  any  fractional  part  of  the  one 
solution  will  neutralize  this  same  fractional  part  of  the 
second  solution. 

In  the  neutralization  of  this  same  base  by  a  dibasic 
acid,  such  as  sulphuric  acid,  the  following  equation  comes 
into  consideration: 

2  Na-  +  2  OH'  +   2  IT   +  SO/  =2  Na'  +  SO/  +  2  H2O 
2(23)  +2(17.01)  +2(1.01)  +96.07  =  2(23)  +96.07  +  2(18.02) 

From  this  it  is  evident  that  a  molar  solution  of  sulphuric 
acid,  with  98.09  grams  of  the  acid  per  liter,  will  contain 
2.02  grams  of  ionizable  hydrogen,  a  quantity  that  re- 
quires 34.02  grams  (2  X  17.01)  of  ionizable  hydroxyl  for 
its  complete  neutralization.  This  quantity  of  hydroxyl 
is  furnished,  as  the  equation  indicates,  through  the  use 


108  CHEMICAL   CALCULATIONS 

of  two  gram-molecular  weights  of  sodium  hydroxide.  If 
we  were  dealing  with  molar  solutions  of  these  substances, 
two  volumes  of  the  sodium  hydroxide  solution  would  be 
required  for  the  neutralization  of  one  volume  of  a  molar 
sulphuric  acid  solution. 

By  reason  of  this  variation  in  the  number  of  ionizable 
hydroxyl  and  hydrogen  groups  in  the  various  substances, 
it  is  found  advisable  to  base  our  standard  solutions  upon 
the  exact  number  of  these  univalent  groups  which  they 
contain,  rather  than  upon  the  entire  molecular  weight  of 
the  substance  itself.  A  solution  which  contains  in  one 
liter  exactly  1.01  grams  of  ionizable  hydrogen  is  taken 
as  the  standard  for  acids,  while  that  which  contains  in 
one  liter  exactly  17.01  grams  of  ionizable  hydroxyl  is 
taken  as  the  standard  for  bases.  These  two  solutions 
are,  volume  for  volume,  always  equivalent  and  may  be 
termed  Equivalent  Normal  Solutions  or,  as  is  more  gen- 
erally the  case,  Normal  Solutions. 

The  molar  solution  of  sodium  hydroxide  is  identical,  of 
course,  with  its  normal  solution.  The  molar  solution  of 
sulphuric  acid  contains  twice  as  much  ionizable  hydrogen 
as  is  required  for  its  normal  solution,  —  a  fact  indicated 
by  the  use  of  two  volumes  of  tlie  molar  sodium  hydroxide 
solution  above  to  neutralize  only  one  volume  of  this  acid. 
We  are  therefore  required  to  dissolve  one-half  of  the 
gram-molecular  weight  (98.09)  of  sulphuric  acid,  or 
49.04  grams,  in  water  and  bring  this  to  one  liter  in  order 
to  obtain  its  true  normal  solution. 

In  the  same  manner  a  base  such  as  barium  hydroxide, 
Ba(OH)2,  with  a  molecular  weight  of  171.39,  will  con- 
tain in  its  molar  solution  171.39  grams  of  substance  of 
which  34.02  grams  is  ionizable  hydroxyl.  A  definite 
volume  of  this  molar  solution  would  neutralize  two 
volumes  of  a  normal  hydrochloric  acid  solution;  conse- 


NORMAL   SOLUTIONS  109 

quently  to  obtain  a  solution  of  17.01  grams  of  hydroxyl 
to  the  liter  (a  normal  solution)  we  should  need  to  dis- 
solve 171.39/2  grams  of  the  substance  in  a  liter  of  solu- 
tion. A  solution  of  this  concentration  is  here  unattain- 
able, as  the  solubility  falls  below  the  value  required.  In 
such  cases  various  degrees  of  dilution  are  used,  as  will  be 
indicated  below. 

Solutions  that  furnish  neither  hydrogen-  nor  hydrox- 
ide-ion are  considered  normal  when  they  contain,  per 
liter,  an  equivalent  of  1.01  grams  of  hydrogen  or  17.01 
grams  of  hydroxyl.  Triis  signifies  that  in  the  reactions 
in  wRTch  they  are  concerned  they  will  have,  per  liter,  the 
power  of  combining  with  or  displacing,  either  directly  or 
indirectly,  these  proportional  amounts  of  hydrogen  or 
hydroxyl.  Thus,  in  the  action  of  hydrochloric  acid  upon 
sodium  carbonate: 

Na2C03  +  2  HC1  =  2  NaCl  +  H2O  +  C02, 

it  is  observed  that  one  gram-molecular  weight  of  the  car- 
bonate brings  into  the  reaction  2.02  grams  of  hydrogen; 
consequently  one-half  of  its  gram-molecular  weight  (106), 
or  53  grams  of  sodium  carbonate,  is  required  in  1  liter 
of  its  normal  solution. 

Normality  Factors.  —  For  chemical  purposes  it  is  not 
necessary  to  bring  every  solution  to  the  same  standard  of 
concentration,  —  the  normal  solution.  The  variations 
from  the  true  normality  may  be  readily  expressed  by 
fractions,  or  factors,  which  designate  at  once  the  actual 
concentration  of  the  solutions  in  terms  of  the  normal. 
Thus  a  molar  solution  of  sulphuric  acid  contains  twice 
what  a  normal  solution  should  contain.  Its  normality, 
therefore,  is  2,  and  is  expressed  as  2  N.  A  solution  con- 
taining 0.365  gram  of  hydrogen  chloride  per  100  c.c.  would 
contain  3.65  grams  per  liter.  This  is  1/10  of  what  a 


110  CHEMICAL    CALCULATIONS 

normal  solution  contains;  hence  the  solution  is  N/10  or 
0.1  N,  (deci-normal). 

By  another  method  of  procedure,  this  solution,  contain- 
ing 0.365  gram  of  hydrogen  chloride  per  100  c.c.,  may  be 
compared  directly  with  the  amount  required  in  a  liter  of 
a  normal  solution,  namely,  36.5  grams:  — 36.5:  0.365  = 
1  :  x.  The  factor  (x)  is  here  0.01,  hence  this  weight  (0.365 
gram)  of  hydrogen  chloride  would  be  contained  in  1/100 
of  1000  c.c.,  or  10  c.c.  of  the  normal  acid.  As  it  actually 
occurs  in  100  c.c.,  then  the  solution  in  question  is  10/100  N 
or  N/10;  that  is,  100  c.c.  of  this  solution  is  necessarily 
equivalent  to  10  c.c.  of  the  normal  solution. 

By  reason  of  the  equivalence  between  equal  volumes  of 
normal  solutions  we  can  readily  calculate  the  normality  of 
any  solution  if  we  are  given  the  normality  of  that  solution 
which  is  to  be  titrated  against  it. 

Example  48.  —  100  c.c.  of  N/2  sodium  hydroxide  solution 
were  required  in  the  neutralization  of  400  c.c.  of  an  unknown 
acid  solution.  Calculate  the  normality  of  this  acid? 

Here,  of  course,  100  c.c.  of  N/2  solution  is  the  equivalent 
of  50  c.c.  of  a  normal  solution;  i.e.,  in  100  c.c.  of  N/2 
sodium  hydroxide  solution  we  have  100/1000  or  1/10  of 
17.01/2  grams,  or  17.01 /20» gram,  of  hydroxyl,  which  is 
exactly  the  amount  contained  in  1/20  of  a  liter  (50  c.c.) 
of  a  normal  solution  containing  17.01  grams  per  liter.  In 
order  to  find  the  normality  of  the  unknown  solution  it  will 
be  necessary  to  get  some  expression  for  it  in  terms  of  the 
known  or  normal  solution.  For  example,  400  c.c.  of  this 
unknown  solution  neutralize  50  c.c.  of  the  normal.  By 
reduction,  1  c.c.  neutralizes  1/8  c.c.  of  the  normal.  The 
relative  volumes  are  equivalent;  hence  1  liter  of  the 
unknown  solution  must  neutralize,  and  possess  an  equiva- 
lent amount  of  substance  to,  that  contained  in  one-eighth 


NORMAL   SOLUTIONS  111 

of  a  liter  of  normal  sodium  hydroxide  solution.  It  is, 
therefore,  but  1/8  N,  a  fact  indicated  at  once  by  the 
factor  obtained  for  the  equivalent  of  that  unit  value  1  c.c. 
above.  If  we  had  taken  as  the  unit  value  1  c.c.  of  the 
known  solution,  we  should  have  obtained  8  as  a  factor 
denoting  the  number  of  cubic  centimeters  of  the  unknown 
acid  solution  equivalent  in  value  to  1  c.c.  of  the  known. 
This  comes  to  the  same  end  and  indicates  the  strength  of 
1  c.c.  of  the  unknown  solution  as  1/8  that  of  the  normal. 
Calculation  of  Normality  by  Simple  Proportion.  —  A 
very  simple  method  for  calculations  of  this  sort  rests  upon 
the  consideration  of  the  proportionality  which  exists 
between  the  normality  factors  and  the  volumes  for  these 
equivalent  solutions. 

Example  49.  — 400  c.c.  of  N/8  acid  solution  neutralized  100 
c.c.  of  an  unknown  alkaline  solution.  Calculate  the  normality 
of  the  alkali. 

Now  the  volume  of  a  solution  when  multiplied  by  its 
normality  factor  gives,  as  we  have  seen,  the  equivalent 
volume  in  terms  of  its  normal  solution.  As  all  of  these 
solutions  are  balanced  or  titrated  to  an  end-point  which 
signifies  that  equivalent  quantities  of  the  various  sub- 
stances are  present,  we  may  at  once  place  the  two  expres- 
sions for  the  two  solutions  as  equal  to  each  other.  Thus : 
400  X  1/8  =  1QO  X  x. 

i  (y  "&   \1   t    Stf**1^  *"  ^PA   */"      O 

All  of  this  is  in  exact  atccord  with  our  premises  which  make 
it  necessary  for  equal  volumes  to  neutralize  equal  volumes 
when  an  equivalent  amount  of  substance  is  present  in  each. 
By  separating  these  terms  of  the  equation  into  means  and 
extremes  of  a  simple  proportion  (for  example  by  dividing 
through  by  the  quantity  100  X  1/8)  we  obtain 

400        x_ 

100  ~  1/8* 


112  CHEMICAL   CALCULATIONS 

This  is  synonymous  with  saying  that  the  normality  factors 
of  the  two  equivalent  solutions  stand  to  each  other  in  an 
inverse  proportion  to  the  corresponding  volumes  required 
in  the  titration: 

100  :  400  =  1/8  :  x. 

By  calculation  x  is  found  to  be  1/2,  i.e.,  the  alkaline  solu- 
tion is  0.5  normal. 

Calculation  of  the  Weights  of  Substances  Present  in 
Standard  Solutions.  —  When  the  normality  of  an  unknown 
solution  has  been  determined  it  is  often  desirable  to  find 
the  exact  amount  of  substance  in  a  given  volume  of  this 
solution.  For  example,  in  the  last  paragraph  the  normal- 
ity of  the  alkaline  solution  was  found  to  be  N/2.  If  it  is 
now  desired  to  learn  the  amount  of  sodium  hydroxide 
actually  present  in  the  100  c.c.  of  solution,  we  need  only 
take  the  proportional  amount  of  sodium  hydroxide  in 
a  liter  of  N/2  solution  (20  grams)  as  is  indicated  by  the 
fractional  part  which  this  volume  is  of  1000,  —  100/1000 
or  1/10,  i.e.,  1/10  of  20  grams,  or  2  grams.  By  simple 
proportion  a  comparison  of  the  volume  relations  with  the 
corresponding  weights  of  the  substances  gives  the  follow- 
ing: 

1000  : 


Standardization  of  Solutions  by  Gravimetric  Means.  — 
For  our  standard  solutions,  it  is  usually  customary  to 
dissolve  a  certain  calculated  amount  of  substance  in  water 
and  bring  the  volume  up  to  the  desired  mark  by  the 
gradual  addition  of  more  water.  As  these  solutions  may 
vary  slightly  from  the  true  values,  it  is  always  desirable 
to  standardize  them  by  purely  chemical  means,  such  as 
titration  against  certain  accurately  prepared  solutions,  or 
if  possible  by  the  formation  of  precipitates  in  a  known 
volume  of  their  solution.  These  precipitates  when  dried 


NORMAL   SOLUTIONS  113 

and  weighed  serve  as  a  means  for  the  calculation  of  the 
exact  normality. 

Example  50.  —  100  c.c.  of  a  hydrochloric  acid  solution;  made 
up  approximately  to  1/5  N,  gave,  when  treated  with  a  slight 
excess  of  a  silver  nitrate  solution,  3.22  grams  of  silver  chloride. 
What  is  the  normality  of  the  acid? 

From  the  equation : 

AgN03  +  HC1  =  AgCl  +  HN03, 

the  proportionality  between  all  amounts  of  hydrogen 
chloride  and  silver  chloride  are  indicated  by  the  ratio : 

AgCl  :    HC1 

(107.88  +  35.46)  :    36.47. 

From  this  ratio  the  weight  of  hydrogen  chloride,  0.8195 
gram,  corresponding  to  the  weight  of  silver  chloride,  3.22 
grams,  is  easily  calculated : 

AgCl    :  HC1     -  3.22  :  x 
143.34  :  36.47  -  3.22  :  0.8195. 

This  weight  of  hydrogen  chloride  is  found  present  in  100  c.c. 
of  solution.  In  one  liter  we  shall  have  8.195  grams,  whereas 
we  should  have  36.47  grams  if  it  were  a  normal  solution. 
The  fraction  8.195/36.47  represents  then  the  normality, 
expressed  decimally  as  0.2247  N,  and  gives  to  the  solution 
a  value  somewhat  higher  than  that  estimated,  (N/5). 

Standardization  of  Solutions  by  Volumetric  Means.  —  In 
place  of  the  method  of  precipitates  another  very  instruct- 
ive method  is  applicable  in  standardization;  chiefly  with 
acids.  This  consists  in  measuring  the  volume  of  a  gas 
evolved  by  the  action  of  some  substance  upon  a  known 
volume  of  the  acid  solution. 

Example  51.  — 250  c.c.  of  an  acid  solution  gave,  when  acted 
upon  by  zinc,  560  c.c.  of  hydrogen  (calculated  at  standard  con- 
ditions of  temperature  and  pressure).  What  is  the  normality 
of  the  acid. 


114  CHEMICAL   CALCULATIONS 

560  c.c.  of  hydrogen  from  250  c.c.  of  solution  would 
mean,  of  course,  2240  c.c.  from  one  liter  of  the  solution. 
By  definition  a  normal  solution  of  an  acid  is  one  that 
contains  1.01  grams  of  ionizable  hydrogen  per  liter.  The 
gram-molecular  weight  (2.02)  of  hydrogen  occupies  a 
volume  of  22,400  c.c.  at  standard  conditions.  The 
volume  occupied  by  1.01  grams  therefore  will  be  just 
one-half  this,  or  11,200  c.c.  Accordingly  every  liter  of  a 
normal  acid  solution  must  contain  that  weight  of  hydro- 
gen which  when  set  free  will  occupy  11,200  c.c.  under  the 
standard  conditions  of  temperature  and  pressure. 

It  is  only  a  simple  step  to  calculate  the  volume  of 
hydrogen  per  liter  when  we  have  given  the  volume  for 
any  fraction  of  a  liter.  In  the  problem  above,  250  c.c. 
of  solution  evolved  560  c.c.  of  hydrogen;  consequently 
1  liter  will  evolve  2240  c.c.  of  this  gas:  250  : 1000  =  560  :  x. 
As  a  liter  of  normal  acid  should  give  11,200  c.c.,  the 
solution  in  question  is  less  than  normal,  and  in  accord- 
ance with  the  ratio 

11,200  :  2240  =  1  :  x,  or  x  =  1/5, 

i.e.,  the  acid  is  0.2  N.  In  other  words,  the  normality  is 
expressed  by  the  fraction,  or  factor,  which  the  volume  of 
hydrogen  evolved  per  liter  makes  with  the  total  volume 
of  hydrogen  that  can  be  evolved  from  a  liter  of  the  normal 
acid. 

If  the  hydrogen  is  measured  at  room  temperature  and 
pressure,  it  is  only  necessary  to  calculate  the  volume  it 
would  occupy  at  the  standard  conditions  and  thus  make 
possible  the  comparison  between  this  and  the  standard 
volume  per  liter.  If  this  is  not  done,  then  the  standard 
volume  per  liter  (11,200  c.c.)  must  be  calculated  to  the 
conditions  of  the  experiment  under,  which  the  gas  is 
measured. 


NORMAL   SOLUTIONS  115 

Similarly,  the  evolution  of  other  gases  by  chemical 
action,  from  known  volumes  of  solutions,  may  serve  for 
the  estimation  of  the  normalities  of  these  solutions.  For 
example,  a  solution  of  sodium  carbonate  may  be  treated 
with  an  acid  and  the  carbon  dioxide  set  free  measured. 
The  following  equation  is  here  involved: 

Na2CO3  +  H2SO4   =  Na2SO4   +  C02  +  H2O. 

The  direct  proportionality  between  the  reaction-quan- 
tities concerned  is  expressed  by  the  ratio  CO2/Na2CO3 
or  44/106.  This  shows  that  for  every  gram-molecular 
weight  of  carbon  dioxide  (44)  we  must  estimate  the  pres- 
ence of  one  gram-molecular  weight  of  sodium  carbon- 
ate (106).  Now  a  normal  solution  of  sodium  carbonate 
contains  only  one-half  of  the  gram-molecular  weight  in 
1  liter;  a  fact  readily  determined  by  its  titration  with  a 
normal  solution  of  any  acid.  This  is  indicated  in  the 
equation  above,  wherein  we  note  that  2.02  grams  of 
ionizable  hydrogen  (2  H'  +  SO/')  are  required  for  the 
complete  action,  and  consequently  only  one-half  of  the 
gram-molecular  weight  of  the  carbonate,  53  grams,  can 
be  equivalent  to  1.01  grams  of  hydrogen.  The  amount 
of  carbon  dioxide  evolved  from  one  gram-molecular 
weight  of  the  carbonate  is  one  gram-molecular  weight, 
or  a  volume  of  22,400  c.c.;  hence  from  a  normal  solu- 
tion with  one-half  the  gram-molecular  weight  of  car- 
bonate, we  should  have,  per  liter,  just  11,200  c.c.  of 
this  gas. 

When  the  volume  of  carbon  dioxide  evolved  from  a 
definite  volume  of  solution  is  known  we  need  only  to 
calculate  the  volume  of  gas  evolved,  per  liter,  and  com- 
pare this  volume  at  standard  conditions  with  the  stand- 
ard volume,  11,200  c.c.  The  ratio  to  this  value  gives  the 
ratio  to  unit  normality. 


116  CHEMICAL   CALCULATIONS 

Example  52.  —  What  is  the  normality  of  a  sodium  carbonate 
solution,  125  c.c.  of  which  evolved  350  c.c.  of  carbon  dioxide 
(at  standard  conditions)  when  treated  with  an  excess  of  acid? 

Since  125  c.c.  of  the  solution  gave  350  c.c.  of  the  gas, 
1000  c.c.  will  give  2800  c.c.  as  determined  from  the  pro- 
portion 125  :  1000  =  350  :  x.  A  normal  solution  should 
evolve  11,200  c.c.  of  this  gas;  consequently  the  solution 
in  question  is  only  2800/11,200  or  1/4  N. 

The  evolution  of  carbon  dioxide  in  the  equation  above 
may  serve  equally  well  in  determining  the  normality  of 
the  sulphuric  acid  used.  Each  liter  of  normal  sulphuric 
acid  will  liberate  11,200  c.c.  of  the  gas.  The  ratio  of 
comparison,  therefore,  is  carried  out  just  as  described  in 
the  previous  paragraph. 

Comparison  of  Solutions  with  Standard.  —  When  a 
solution  is  once  standardized  other  solutions  may  be 
standardized  by  comparison  with  it  either  directly  or  j 
indirectly.  In  the  case  of  a  second  acid  solution  it  is, 
necessary  to  ascertain  what  volumes  of  both  this  acid 
and  our  standard  acid  are  required  for  the  neutralization 
of  equal  volumes  of  some  alkaline  solution.  These  two 
equivalent  volumes  are  then  compared  just  as  in  the 
preceding  examples. 

Example  68.  —  10  c.c.  of  1.5  N  hydrochloric  acid  neutralized 
40  c.c.  of  an  alkaline  solution.  50  c.c.  of  an  unknown  sulphuric 
acid  solution  neutralized  80  c.c.  of  this  same  alkaline  solution. 
Calculate  the  normality  of  the  sulphuric  acid. 

50  c.c.  of  the  sulphuric  acid  neutralized  80  c.c.  of  the 
alkali;  25  c.c.  of  the  acid  would  neutralize  40  c.c.  of  the 
alkali.  This  is  the  same  volume  of  alkali  neutralized  by 
10  c.c.  of  1.5  N  hydrochloric  acid;  hence  these  two 
volumes  of  the  acid  solutions  must  be  equivalent: 

10  c.c.  X  1.5  N  HC1  =  25  c.c.  X  (x)  N  H2SO4. 


NORMAL   SOLUTIONS  117 

By  proportion : 

25  :  10  =  1.5  :  xt  or  x  =  3/5. 

Hence  the  normality  t>f  the  sulphuric  acid  is  found  to  be 
3/5,  i.e.,  0.6  N. 

Adjustment    of    Solutions    to    a    Desired    Standard.  — 

When  a  solution  has  been  standardized  and  found  to 
vary  somewhat  from  the  estimated  normality,  it  is  cus- 
tomary to  calculate  the  amount  of  substance  (if  too 
dilute)  or  water  (if  too  concentrated)  that  will  bring  the 
solution  to  the  desired  normality.  The  latter,  which  con- 
stitutes the  more  simple  case,  is  illustrated  in  the  follow- 
ing example: 

Example  64.  —  A  solution  of  hydrochloric  acid  was  found  to 
have  a  normality  of  1.05.  What  volume  of  water  must  be 
added  to  400  c.c.  of  this  solution  to  make  it  exactly  normal? 

1  c.c.  of  1.05  N  hydrochloric  acid  is  equivalent  by 
definition  to  1.05  c.c.  of  a  normal  hydrochloric  acid 
solution.  Therefore,  to  make  this  more  concentrated 
acid  normal,  we  need  only  add  water  until  1.05  c.c.  of 
the  diluted  solution  will  be  exactly  equivalent  to  1.05  c.c. 
of  a  normal  solution;  in  other  words, 

1.05  c.c.  —  1  c.c.  =  0.05  c.c., 

or  that  volume  of  water  required  per  cubic  centimeter  of 
the  1.05  N  acid.  400  c.c.  will  require  400  X  0.05  c.c.,  or 
20  c.c.  Therefore,  when  400  c.c.  of  this  1.05  N  acid  are 
diluted  with  20  c.c.  of  water,  the  final  420  c.c.  will  be  just 
normal. 

By  comparison  of  the  actual  weights  of  hydrogen  chlo- 
ride in  equal  volumes  of  these  acids  and  the  direct  pro- 
portionality between  these  weights  and  the  corresponding 
volumes  which  are  equivalent,  volume  for  volume,  we  may 
calculate  the  amount  of  dilution  necessary  to  bring  any 


118  CHEMICAL  CALCULATIONS 

known  solution  to  a  desired  normality.  Thus  in  1  c.c.  of 
normal  hydrochloric  acid  there  is  0.0365  gram  of  hydro- 
gen chloride,  and  in  1  c.c.  of  1.05  N  hydrochloric  acid 
there  is  1.05  X  0.0365  gram  of  hydrogen  chloride.  There- 
fore 

0.0365':  (1.05  X  0.0365)  =  400  :  x,  or  x  =  420. 

This  gives  the  volume  to  which  400  c.c.  of  1.05  N  acid 
must  be  diluted. 

When  a  solution  is  found  too  dilute,  a  similar  calcula- 
tion will  give  the  amount  of  water  that  should  be  removed 
from  the  volume  in  question.  Without  resorting  to  this 
procedure,  it  is  found  better  to  add  to  the  entire  volume 
that  weight  of  substance  (usually  a  well-defined  salt) 
necessary  to  make  with  this  excess  of  water  a  solution  of 
the  desired  normality.  Any  change  in  volume  due  to 
process  of  solution  of  the  salt  may  be  neglected. 

The  calculation  of  results  through  reactions  which  in- 
volve solutions  of  definite  concentration  follows  the  general 
outlines  presented  in  Chapter  IX.  The  amount  of  any 
substance  present  in  a  required  volume  of  a  standard 
solution  is  to  be  considered,  of  course,  with  reference  to 
its  corresponding  reaction-quantity. 

The  method  of  determining  the  amount  of  a  substance 
necessary  to  complete  a  given  reaction  with  a  known 
amount  of  some  other  substance  contained  in  a  definite 
volume  of  solution  is  called  "  Volumetric  Analysis."  When 
no  reference  is  made  to  volume  relations,  but  calculations 
are  made  upon  the  weighed  quantities  which  come  under 
consideration,  we  have  the  more  common  "  Gravimetric 
Analysis."  These  two  form  the  basis  of  work  in  quanti- 
tative chemical  analysis. 


NORMAL   SOLUTIONS  119 


PROBLEMS. 

185.  400  c.c.  of  N/4  potassium   hydroxide  solution  were 
required  for  the  neutralization  of  600  c.c.  of  an  unknown  acid 
solution.     Calculate  the  normality  of  this  acid  solution. 

Ans.  N/6. 

186.  500  c.c.  of  N/10  acid  solution  were  required  for  the 
neutralization  of  25  c.c.   of  a  solution  of  sodium  hydroxide. 
Calculate  the  normality  of  this  latter  solution.  Ans.  2  N. 

187.  220  c.c.  of  N/20  acid  solution  were  required  for  titra- 
tion  with  124  c.c.  of  a  solution  of  barium  hydroxide.     Calcu- 
late the  normality  factor  of  the  barium  hydroxide  solution. 

Ans.  .0887  N. 

188.  What  volume  of  N/10  acid  solution  will  be  required  in 
the  titration  of  440  c.c.  of  N/4  sodium  hydroxide  solution? 

Ans.  1100  c.c. 

NOTE:  —  The  method  by  simple  proportion,  with   one  of  the 
volumes  as  the  unknown  term,  will  be  found  to  serve  well  in  such 


189.  What  volume  of  N/6  alkaline  solution  will  be  required 
in  the  titration  of  254  c.c.  of  N/10  acid  solution? 

Ans.  152.4  c.c. 

190.  Calculate  the  weight  of  hydrogen  chloride  present  in 
400  c.c.  of  a  hydrochloric  acid  solution  which  required  320  c.c. 
of  N/4  alkaline  solution  for  titration.  Ans.  2.918  grams. 

191.  Calculate  the   weight    of    sulphuric    acid    present   in 
150  c.c.  of  a  solution  which  required  48.1  c.c.  of  0.78  N  alkali 
for  titration.  Ans.  1.84  grams. 

193.  An  excess  of  silver  nitrate  solution  was  added  to  350 
c.c.  of  a  solution  of  hydrochloric  acid.  The  precipitate  of  silver 
chloride  weighed  7.54  grams.  Calculate  the  normality  of  the 
acid.  Ans.  0.1503N. 

193.  A  slight  excess  of  barium  chloride  solution  was  added 
to  400  c.c.  of  a  solution  of  sulphuric  acid.  From  the  weight  of 
barium  sulphate,  BaSO4,  precipitated,  4.12  grams,  calculate 
the  normality  of  the  acid.  Ans.  0.0882  N. 


120  CHEMICAL   CALCULATIONS 

194.  600  c.c.  of  a  sulphuric  acid  solution,  when  acted  upon 
by  an  excess  of  zinc,  evolved  1242  c.c.  of  hydrogen  (at  standard 
conditions).     Calculate  the  normality  of  the  acid. 

Ans.  0.1848N. 

195.  440  c.c.  of  an  acid  solution,  when  acted  upon  by  an 
excess  of  zinc,  evolved  2430  c.c.  of  hydrogen,  measured  over 
water  at  21°  and  747.5   mm.   pressure.      Calculate   the    nor- 
mality of  the  acid.  Ans.  0.4408  N. 

196.  Calculate  the  normality  of  an  acid  solution,  600  c.c. 
of  which,  when  acted  upon  by  an  excess  of  sodium  carbonate, 
evolved  2100  c.c.   of  carbon  dioxide   (calculated  to  standard 
conditions).  Ans.  0.3125  N. 

197.  Calculate  the    normality  of   a  solution  of   potassium 
carbonate,  200  c.c.  of  which,  when  treated  with  an  excess  of 
acid,    evolved    4502    c.c.    of   carbon    dioxide    (calculated    to 
standard  conditions).  Ans.  2.01  N. 

198.  An  excess  of  iron  sulphide,  FeS,  was  added  to  500  c.c. 
of  a  solution  of  sulphuric  acid.     The  volume  of  hydrogen  sul- 
phide  set  free   measured  4640  c.c.    (at  standard   conditions). 
Calculate  the  normality  of  the  acid.  Ans.  0.8285  N. 

199.  An  excess  of  sodium  sulphite,  Na2S03,  was  added  to 
400  c.c.   of  a  solution  of  hydrochloric  acid.     The  volume  of 
sulphur  dioxide  set  free  measured  5600  c.c.  (at  standard  con- 
ditions).    Calculate  the  normality  of  the  acid.      Ans.  1.25  N. 

A 

200.  1400  c.c.  of  ammonia  (calculated  to  standard  condi- 
tions) were  passed  into  500  c.c.  of  N/2  hydrochloric  acid  solu- 
tion.    Calculate  the  normality  of  the  hydrochloric  acid  still 
present.  Ana.  3/8  N. 

/|  201.  210  c.c.  of  carbon  dioxide  (at  standard  conditions) 
were  passed  into  250  c.c.  of  N/10  barium  hydroxide  solution. 
Calculate  the  normality  of  the  barium  hydroxide  solution  still 
present.  Ans.  N/40. 

202.  50  c.c.  of  N/5  hydrochloric  acid  solution  neutralized 
40  c.c.  of  an  unknown  alkaline  solution.  300  c.c.  of  a  sul- 
phuric acid  solution  neutralized  60  c.c.  of  this  same  alkaline 
solution.  Calculate  the  normality  of  the  sulphuric  acid. 

Ans.  0.05  N. 


NORMAL   SOLUTIONS  121 

203.  200  c.c.  of  a  barium  hydroxide  solution  were  required 
in  the  titration  of  40  c.c.  of  an  acid  solution.     100  c.c.  of  this 
acid  solution  exactly  neutralized  80  c.c.  of  N/2  alkaline  solu- 
tion.    Calculate  the  normality  of  the  barium  hydroxide  solu- 
tion. Ans.  0.08  N. 

204.  A  solution  of  hydrochloric  acid  is  desired  to  be  made 
exactly  normal.      40   c.c.  of  the  solution  neutralized   50  c.c. 
of  0.84  N  sodium  hydroxide  solution.     Calculate  the  volume  of 
water  that  must  be  added  per  100  c.c.  of  the  acid  solution. 

Ans.  5  c.c. 

205.  A  solution  of  sodium  hydroxide  is  desired  to  be  made 
exactly  0.5  N.     32  c.c.  of  the  solution  at  hand  were  required 
for  the  titration  of  28  c.c.  of  0.8  N  hydrochloric  acid.     Calcu- 
late the  volume  of  water  that  must  be  added  per  100  c.c.  of  the 
alkaline  solution.  Ans.  40  c.c. 

206.  A  solution  of  sodium  carbonate  is  desired  to  be  made 
exactly  0.05  N.     24  c.c,   of  the  solution  at  hand  neutralized 
9.6  c.c.   of  0.12  N  hydrochloric  acid  solution.     Calculate  the 
weight  of  anhydrous  salt,  Na2CO3,  that  must  be  added  per  100 
c.c.  of  solution.  Ans.  0.0106  gram. 

207.  What  weight  of  iron  will  be  required  for  interaction 
with  400  c.c.  of  N/5  hydrochloric  acid?         Ans.  2.234  grams. 

208.  What  weight  of  sodium  carbonate  will  be  required  for 
interaction  with  600  c.c.  of  N/8  sulphuric  acid?     What  volume 
of  carbon  dioxide  (at  standard  conditions)  will  be  evolved? 

Ans.  3.975  grams. 
840  c.c.  CO2. 

209.  What  weight  of  sodium  hydrogen  carbonate,  NaHC08, 
will  be  required  for  interaction  with  600  c.c.  of  N/8  sulphuric 
acid?     What  volume  of  carbon  dioxide  (at  standard  conditions) 
will  be  evolved?  Ans.  6.3  grams. 

1680  c.c.  COa. 

210.  Calculate    the    weight    of    crystallized    oxalic    acid, 
C2H204 .  2  H20,  required  for  a  solution  which  is  to  be  made  up 
to  500  c.c.  in  volume  at  N/2.  Ans.  15.75  grams. 

211.  Calculate  the  volume  of  nitric  oxide,  NO  (at  standard 
conditions)  that  could  be  evolved  by  the  action  of  copper  upon 
1000  c.c.  of  a  7  N  nitric  acid  solution.  Ans.  39,200  c.c. 


CHAPTER  XI. 
COMBINATIONS   BETWEEN   GASES   BY   VOLUME. 

WHEN  expressed  in  grams  the  molecular  weights  of  all 
substances  in  the  state  of  vapor  occupy  a  volume  of 
22,400  c.c.  at  the  standard  conditions  of  temperature  and 
pressure.  In  Chapter  IX  it  was  observed  that  a  molecular 
equation  offered  for  this  reason  an  insight  into  the  volume 
relations  of  the  various  substances  concerned  in  a  given 
reaction. 

Molecular  Volumes.  —  These  volumes  comply  naturally 
with  the  laws  relating  to  gases,  and,  further,  are  subject  to 
the  operation  of  those  properties,  characteristic  of  each 
substance,  which  may  here  be  brought  into  consideration 
through  the  conditions  of  the  experiment.  Thus  the  con- 
densation of  a  gas  or  vapor  to  the  liquid  or  solid  state,  or 
its  solution  in,  or  combination  with,  various  substances 
which  may  be  present,  will  remove  it  completely  from 
further  considerations. 

As  a  simple  illustration  of  these  facts  an  excellent  ex- 
ample is  found  in  the  combination  of  hydrogen  with 
oxygen: 

2  H2  +  O2  =  2  H2O. 

By  molecules:          2       +1=2. 

By  gram-molecular 
volumes : 

2  (22,400)  c.c.  -f  1  (22,400)c.c.  =  2  (22,400)  c.c.  + 

(22,400  c.c.  contraction). 

122 


COMBINATIONS   BETWEEN   GASES   BY   VOLUME        123 

Reduced  through  the 
common  term,  22,400: 

2  vol.  +  1  vol.  =  2  vol.  +  (1  vol.  contr.). 

Each  gram-molecular  weight  may  be  represented  by 
one  volume,  i.e.,  its  gram-molecular  volume.  The  coeffi- 
cients or  integers  which  designate  the  number  of  gram- 
molecular  weights  will  stand  likewise  for  the  number  of 
gram-molecular  volumes  involved  in  the  reaction.  The 
combination  of  gases  by  volume  expressed  thus  by  simple 
numbers  has  been  developed  experimentally  and  is  com- 
prehended in  Gay-Lussac's  well-known  Law  of  Combining 
Volumes. 

In  the  reaction  above  we  observe  that  2  volumes  of 
hydrogen  and  1  volume  of  oxygen  unite  to  form  2  volumes 
of  aqueous  vapor,  the  temperature  of  100°  or  above, 
and  the  observed  pressure,  remaining  constant  through- 
out. This  diminution  in  the  total  volume  of  the  gaseous 
components  when  transformed  into  aqueous  vapor 
has  been  considered  in  the  study  of  Avogadro's  hy- 
pothesis. 

If  the  temperature  falls  below  100°  a  further  contraction 
in  volume  occurs.  Since  this  is  due  to  the  condensation 
of  aqueous  vapor  to  the  liquid  state,  and  increases  accord- 
ingly with  a  lowering  of  temperature,  there  is  left,  eventu- 
ally, no  volume  of  gaseous  product.  The  water  formed 
becomes  associated  with  the  liquid  over  which  the  gases 
are  measured,  and  consequently  drops  out  of  further 
consideration  through  the  equalization  or  adjustment  of 
the  levels  within  and  without  the  vessel  to  bring  all  to 
uniform  pressure.  Of  course  a  small  amount  of  water 
remains  in  the  vapor  state,  even  at  a  low  temperature; 
corrections  for  this  are  made  by  reference  to  a  table  of 
aqueous  vapor  tensions  (Appendix  II). 


124  CHEMICAL   CALCULATIONS 

The  volume  of  the  dry  gaseous  product,  which  in  this 
case  is  theoretically  nil,  may  be  expressed  in  this  manner: 

2  H2  +  O2  =  2  H2O. 

Molecular  volumes:  2       +1     =  2  +  (1  vol.  contr.). 
Molecular  volumes 
below  100°:  2        +1     =  0  +  (3  vol.  contr.). 

The  Relation  of  Molecular  Volume  to  a  Definite  Volume- 
Unit.  —  In  the  study  of  these  volume-changes  in  known 
reactions  it  is  always  necessary  to  determine  the  exact 
relation  which  any  measured  volume  of  vapor,  under 
consideration,  bears  to  the  corresponding  molecular  vol- 
ume representing  it  in  the  molecular  equation  governing 
the  reaction. 

The  coefficients  of  the  quantities  in  a  molecular  equation 
determine  the  ratios  between  the  volumes  of  these  sub- 
stances when  in  the  state  of  vapor.  The  coefficient  unity 
stands,  of  course,  for  the  unit  of  molecular  volume  (22,400 
c.c.),  a  volume  corresponding  to  a  gram-molecular  weight 
of  substance.  Any  fractional  part  of  the  gram-molecular 
weight  of  a  substance  will  occupy,  therefore,  under  stand- 
ard conditions,  a  volume  denoted  by  this  same  fractional 
part  of  the  gram-molecular  volume,  22,400  c.c.  What- 
ever may  be  this  fractional  part  of  the  molecular  quantity 
of  a  substance  under  consideration  in  a  reaction,  all  other 
substances  possible  of  interaction  with  this  one  must  be 
correspondingly  reduced  from  their  own  molecular  quan- 
tities. 

These  coefficients,  therefore,  are  constant  for  any  known 
molecular  equation  and  stand  for  the  ratio  between  the 
volumes  of  the  substances  in  state  of  vapor,  be  they 
gram-molecular  volumes  or  any  definite  fractional  parts 
of  these.  That  volume  which  is  represented  by  the  co- 
efficient unity  in  a  molecular  equation  is  indeed  the  unit 


COMBINATIONS    BETWEEN   GASES   BY   VOLUME        125 

upon  which  all  of  the  other  substances,  in  the  state  of 
vapor,  enter  into  the  reaction.  It  may  be  regarded  as  the 
Volume-Unit  for  the  reaction  given.  If  the  coefficient 
representing  a  certain  substance  in  a  molecular  equation 
is  greater  than  unity,  then,  of  course,  from  the  actual 
volume  of  vapor  of  the  substance  here  concerned,  the 
true  volume-unit  for  the  equation  can  be  derived  only 
by  dividing  this  known  volume  by  the  corresponding 
coefficient. 

Calculation  of  the  Volume-Unit  from  a  Single  Known 
Volume.  —  When  once  the  volume-unit  for  a  reaction  is 
established,  the  actual  volumes  of  the  several  substances 
here  entering  into  combination  are  easily  determined 
from  the  product  of  the  several  coefficients  in  the  molecu- 
lar equation  by  the  volume-unit. 

Example  55.  —  What  volume  of  oxygen  will  be  required  to 
burn  300  c.c.  of  hydrogen  and  what  volume  of  aqueous  vapor 
will  result,  the  temperature  of  100°  and  the  atmospheric 
pressure  remaining  constant  throughout? 

2  H2  +  02  =  2  H20. 
Molecular  volumes : 

2     +1     =  2  +  (1  vol.  contr.). 

By  placing  the  molecular  volume  coefficients  out  by 
themselves,  the  relative  volumes  of  the  several  sub- 
stances are  indicated.  Since  we  can  make  our  calcula- 
tions only  from  the  volumes  actually  given  as  data,  we 
refer  the  volume  of  hydrogen,  300  c.c.,  directly  to  the 
coefficient  2.  From  this  we  derive  the  volume  150  c.c. 
(300  -T-  2)  as  the  value  for  the  coefficient  of  unity,  i.e., 
the  volume-unit  of  the  reaction.  There  remains  now 
only  to  substitute  this  value  for  each  coefficient  through- 
out the  entire  molecular  equation. 


126  CHEMICAL   CALCULATIONS 

2  H2  +  O2  =  2  H2O. 

Molecular  volumes :      2+1    =  2  -f  (1  vol.  contr.). 
Volumes  upon  the  unit  150  c.c.: 

300  c.c.  +  150  c.c.  =  300  c.c  +  (150  c.c.  contr.). 

The  solution  of  the  problem  is  clearly  seen.  150  c.c.  of 
oxygen  are  required  for  the  combustion  of  the  300  c.c.  of 
hydrogen,  while  the  volume  of  aqueous  vapor  that  results 
will  measure  300  c.c. 

Calculation  of  Volume-Unit  from  Known  Mixtures  of 
Gases.  —  Since  the  coefficients  or  integers  represent  the 
ratios  between  the  several  volumes  which  enter  into 
chemical  combination,  it  must  be  understood  that  these 
combinations  take  place  only  in  accordance  with  these 
ratios,  and  that,  if  any  substance  is  present  in  amount  to 
exceed  the  stipulated  volume,  this  excess  must  remain 
unaffected  in  the  reaction;  if  present  in  amount  less 
than  the  stipulated  volume,  the  entire  reaction  must  run 
upon  a  somewhat  smaller  scale,  or  volume-unit,  for  this 
particular  substance;  with  the  result  that  the  other  sub- 
stance or  substances  concerned  will  be  in  excess  of  the 
corresponding  volumes  stipulated  by  the  ratios,  and 
consequently  a  fraction  of  each  of  these  will  remain 
unaffected. 

Example  56.  —  A  mixture  of  250  c.c.  of  hydrogen  and  150  c.c. 
of  oxygen  was  submitted  to  the  action  of  an  electric  spark. 
What  was  the  volume  of  the  product  after  the  explosion,  the 
temperature  100°  and  atmospheric  pressure  constant? 

2  H2  +  02  =  2  H20. 

Molecular  volumes :     2      +1      =  2 +(1  vol.  contr.). 
In  making  hydrogen  the  basis  for  our  calculations,  we 
derive  from  the  volume  250  c.c.  the  value  125  c.c.  as  the 
volume-unit.     Substituting  this   value  for  unity  in  the 
equation  above  we  obtain  the  following: 


COMBINATIONS    BETWEEN    GASES   BY    VOLUME        127 

250  c.c.  +  (125  c.c.  contr.). 


Volume 

250  c.c.  + 

125  c.c. 

relations, 

Volumes 

250  c.c.  + 

150  c.c. 

given, 

Excess  in  j 

25  c.c. 

oxygen,    { 

The  volume  of  aqueous  vapor  formed  is  250  c.c.  The 
volume  of  oxygen,  however,  is  seen  to  be  in  excess  by 
25  c.c.  of  the  volume  actually  required  in  the  reaction; 
hence  the  apparent  volume  of  product  will  be  increased 
by  this  same  amount: 

250  c.c.  +  25  c.c.  =  275  c.c., 

which  is  the  volume  of  gaseous  mixture  after  the  explo- 
sion. 

In  place  of  hydrogen  as  the  basis  for  this  calculation 
of  the  volume-unit,  we  may  now  choose  oxygen.  The 
volume  given,  150  c.c.,  will  then  correspond  to  a  volume- 
unit  in  the  equation;  hence  by  substituting  this  value 
throughout  we  obtain 

2  H2  +  02  =2  H2O 

2+1  =2. 

Volume  relations:   300  c.c.  +  150  c.c.  =  300  c.c. 
Volumes  given:       250  c.c.  +  150  c.c. 

Here  the  volume  of  hydrogen  demanded  by  the  equa- 
tion for  combination  with  150  c.c.  of  oxygen  actually 
exceeds  the  amount  of  hydrogen  at  our  command.  Our 
volume-unit,  therefore,  has  been  placed  too  high.  It  is 
necessary,,  then,  to  reduce  these  high  values  until  the 
calculated  volumes  of  all  the  separate  components  fall 
equal  to  or  under  the  volumes  of  substances  actually 
present.  Such,  of  course,  was  true  in  the  selection  of 
hydrogen,  as  above,  for  the  basis  of  the  calculation. 


128  CHEMICAL   CALCULATIONS 

Calculation  of  True  Volume-Unit  for  a  Reaction.  —  In 

general,  we  determine  a  volume-unit  for  a  reaction  from 
each  of  the  known  volumes  of  substances  present;  this 
of  course  by  dividing  each  volume  by  the  corresponding 
coefficient  which  represents  it  in  the  molecular  equation. 
Upon  comparison  of  the  volume-units  thus  derived,  the 
smallest  will  stand  for  the  true  unit.  In  the  example 
above  the  volume-unit  upon  the  hydrogen  basis  is  125  c.c., 
upon  the  oxygen  basis  it  is  150  c.c;  consequently  the  value 
125  c.c.  alone  fulfills  the  requirements  of  the  reaction. 

Example  57.  — 100  c.c.  of  ammonia  and  90  c.c.  of  oxygen 
were  exploded.  What  was  the  final  volume  of  the  product? 
A  temperature  of  100°  and  atmospheric  pressure  constant. 

The  reaction  proceeds  in  accordance  with  the  equation: 
4  NH3  +  3  O2  =  2  N2  +  6  H20. 

Molecular  f 

,  {4+3=2+       6  —  (1  vol.  expan.). 

volumes :    ( 

From  the  volume  of  ammonia: 

100/4  =  25  c.c.  =  volume-unit. 
From  the  volume  of  oxygen: 

90/3  =  30  c.c.  =  volume-unit. 

The  smaller  value,  25  c.c.,  must  be,  therefore,  the 
volume-unit  for  this  reaction,  while  the  oxygen  will  be 
found  slightly  in  excess  of  the  required  amount. 

Substituting  the  value  25  c.c.  for  unity  throughout,  we 
obtain 

4  NH3  +  3  O2  =  2  N2  +  6  H2O 

4+3        =2     +  6- (1vol.  expan.). 
Volume 


100  c.c.  +  75  c.c.  =50  c.c. +  150  c.c. 
100  c.c.  +  90  c.c. 


relations : 

Volumes 

given: 

Excess:  15  c.c.  oxygen. 


COMBINATIONS   BETWEEN   GASES   BY   VOLUME        129 

Therefore  the  volume  of  product  will  be  equal  to 
50  c.c.  +  150  c.c.,  or  200  c.c.,  plus  the  excess  of  oxygen, 
15  c.c.,  remaining  unacted  upon,  or  215  c.c.  in  all. 

The  determination  of  this  volume-unit  in  combinations 
between  gases  is  at  the  basis  of  all  these  calculations.  In 
examples  where  the  volume  of  only  one  substance  is  given 
it  is  often  difficult  to  ascertain  the  volumes  of  other  sub- 
stances that  may  enter  into  the  specified  reaction. 

Calculation  of  Volume-Unit  from  Measured  Volume- 
Changes.  —  Where  calculations  are  based  alone  upon  the 
volumes  of  gases  that  enter  into  combination,  the  exact 
proportion  of  each  and  every  component  of  the  mixture 
must  be  known;  otherwise  the  volume-units  for  the  several 
reactions  cannot  be  determined.  If  one  or  more  of  these 
factors  are  unknown,  then  a  study  of  the  various  con- 
tractions and  expansions  in  the  total  volume  of  product 
over  that  of  the  original  mixture  offers  a  direct  method 
for  t-he  solution.  Heretofore  we  have  concerned  ourselves 
only  with  the  components  that  entered  into  a  reaction. 
The  study  of  the  products,  however,  offers  far  greater 
possibilities  for  the  reason  that  the  numerous  condensa- 
tions and  absorptions  serve  to  estimate  the  volumes  of 
the  many  substances  possible  of  formation.  Through 
these  observations,  and  a  study  of  the  equations  involved, 
we  may  deduce  all  of  the  relations  between  the  various 
components  that  entered  into  a  reaction. 

These  conclusions  are  made  possible  by  the  conditions 
of  a  molecular  equation,  wherein  only  molecular  -volumes 
and  multiples  of  these  are  concerned.  An  expansion  or 
contraction  in  total  volume  of  product  over  the  original 
volume  must  be  represented,  therefore,  by  a  gain  or  loss 
in  a  definite  number  of  molecular  volumes.  This  number 
will  be  denoted  by  the  difference  between  the  sum  of  the 
molecular  volumes  on  one  side  of  the  equation  and  the 


130  CHEMICAL  CALCULATIONS 

sum  of  the  molecular  volumes  on  the  other.  If  this 
difference  corresponds  to  a  single  molecular  volume,  i.e., 
a  volume  with  coefficient  unity,  then  the  expansion  or 
contraction  observed  is  in  reality  the  exact  measure  of 
the  volume-unit  for  the  reaction;  whereas,  if  this  difference 
is  greater  or  less  than  unity,  the  change  in  volume  observed 
must  be  reduced,  through  division  by  the  coefficient  for 
this  difference,  when  a  volume  is  obtained  corresponding 
to  a  coefficient  of  unity,  i.e.,  the  volume-unit. 

Example  58.  — 100  c.c.  of  a  sample  of  air  were  mixed  with 
100  c.c.  of  hydrogen  (an  excess)  and  exploded.  After  the  removal 
of  the  aqueous  vapor  by  absorption,  the  volume  of  dry  gaseous 
product  read  140  c.c.  Calculate  the  percentage  of  oxygen  in 
this  sample  of  air,  the  room  temperature  and  pressure  remain- 
ing a  constant  throughout. 

2  H2  +  02  =  2  H20. 

Molecular  volumes  0   .  /1      ,  x 

at  100°:  =2+(lvol.contr.). 

Molecular  volumes 


at  room  tempera- 


2+1=0  -h(3vol.contr.). 


ture: 

If  the  conditions  were  such  that  all  of  the  water  remained 
in  state  of  vapor  (100°),  the  contraction  observed,  and  rep- 
resented by  a  coefficient  of  unity,  would  stand  for  the  con- 
traction of  1  volume-unit  in  the  equation,  and  from  this 
value  the  volume  of  oxygen  present  —  1  volume-unit  — 
would  be  found  equal  in  value  to  the  contraction  itself. 

With  the  conditions  otherwise  and  the  loss  of  aqueous 
vapor  increasing  the  molecular  volume  contraction  by  2 
molecular  volumes  (2  volume-units),  we  note  that  the 
observed  contraction  must  be  due  to  the  loss  of  3  volume- 
units  from  the  side  of  the  products.  The  original  mixture, 
100  c.c.  +  100  c.c.,  or  200  c.c.,  lost  60  c.c.  in  this  reaction 
(200  c.c.  —  140  c.c.);  hence  60  c.c.  represents  the  3  volume- 


COMBINATIONS   BETWEEN   GASES    BY    VOLUME        131 

units,  or  1  volume-unit  will  be  equal  to  20  c.c.  From  the 
equation,  the  oxygen  consumed  is  exactly  1  volume-unit 
and  hence  20  c.c.  This  indicates  that  20  per  cent  of  the 
sample  of  air  was  oxygen. 

Example  59,  —  A  volume  of  ammonia  was  mixed  with  a 
large  excess  of  oxygen  and  exploded.  The  expansion  in  volume 
of  product  over  the  original  mixture  was  50  c.c.  What  was  the 
volume  of  ammonia?  Temperature  of  100°  and  atmospheric 
pressure  a  constant. 

The  equation  for  this  action  has  been  given  before: 

4  NH3  +  3  O2  =  2  N2  +  6  H20. 
Molecular  f         4      +3     =     2+      6. 
volumes:  {  7  vol.  =  8  vol.  —  (1  vol.  expan.). 

In  this  case  the  expansion  is  seen  to  be  exactly  equal  to 
a  unit  coefficient  in  volume;  consequently  50  c.c.  is  the 
volume-unit  for  the  reaction.  By  substituting  this  value 
in  the  molecular  equation,  the  volume  of  ammonia  (4  vol- 
ume-units), which  in  its  combustion  can  produce  this 
expansion  of  50  c.c.,  is  found  equal  to  4  X  50  c.c.  or 
200  c.c. 

From  a  study  of  the  volume  of  products,  and  the  alter- 
ations in  this  volume  through  elimination  of  certain  of 
the  substances,  we  are  able  to  determine  the  actual  com- 
position of  unknown  mixtures  of  gases. 

Example  60.  —  200  c.c.  of  a  mixture  of  nitrogen  and  methane 
were  exploded  with  400  c.c.  of  oxygen.  The  volume  of  the  dry 
gaseous  product  measured  500  c.c.  What  was  the  percentage 
of  methane  in  the  original  mixture?  The  room  temperature  and 
pressure  a  constant. 

CH4  +  202  =  C02  +   2  H20. 
Molecular  volumes :       1+2=1+2 
Molecular  volumes, 

aqueous  vapor  1     +2     =    1     +  (2  vol.  contr.). 

removed: 


132  CHEMICAL   CALCULATIONS 

Here  no  change  in  volume  results  from  the  explosion  when 
the  temperature  is  100°  or  over.  When  the  temperature 
falls  and  the  aqueous  vapor  is  condensed  we  have  a  new 
relation  between  the  volumes  of  product  and  original 
mixture  such  that  a  contraction  corresponding  to  2  volume- 
units  occurs.  In  the  example  the  actual  contraction  due 
to  this  removal  of  the  water  is  100  c.c.,  (200  +400  -  500). 
From  this  we  derive  the  value  of  a  single  volume-unit 
as  equal  to  50  c.c.  By  reference  then  to  the  volume 
equation  above,  1  volume-unit,  i.e.,  50  c.c.,  represents 
the  amount  of  methane  present  in  the  original  mixture, 
of  which  it  constitutes  25  per  cent  (by  volume).  The 
nitrogen  is  considered  as  without  action. 

In  practice  the  removal  of  aqueous  vapor  is  usually  not 
carried  out.  Since  at  room  temperatures  the  greater  part 
of  this  vapor  condenses  to  the  liquid  state,  the  partial 
pressure  of  this  vapor  at  the  observed  conditions  gives  at 
once  a  means  of  calculating  the  volume  of  dry  gaseous 
product.  Some  definite  conditions  of  temperature  and 
pressure  are  of  course  taken  as  a  basis.  In  this  manner 
a  direct  comparison  between  the  volume  of  product  and 
that  of  the  original  mixture  can  be  readily  made. 

Example  61.  —  A  mixture  of  200  c.c.  of  ethylene  and  800  c.c. 
of  oxygen  (an  excess),  at  24°  and  756.5  mm.  pressure,  was 
exploded.  After  the  explosion  the  volume  of  product  read 
649.1  c.c.,  at  27°  and  this  same  pressure.  Calculate  the  per- 
centage purity  of  the  ethylene  by  volume. 

The  reaction  takes  place  according  to  the  equation : 

C2H4  +  3O2  =  2C02+2H20. 
Molecular  volumes:        1+3     =     2+      2 


Molecular  volumes, 
aqueous  vapor 
removed: 


1+3      =      2     +  (2  vol.contr.) 


COMBINATIONS   BETWEEN   GASES   BY   VOLUME       133 

When  allowance  is  made  in  this  final  volume  of  product 
for  the  presence  of  aqueous  vapor,  which  at  27°  has  a 
tension  of  26.5  mm.,  we  have  only  to  calculate  the  volume 
which  the  gaseous  product  will  occupy  at  the  original  con- 
ditions, 24°  and  756.5  mm.  pressure,  the  basis  thus  selected 
for  comparison.  The  expression  is 


V  =  620  c.c. 

The  original  mixture,  200  c.c.  +  800  c.c.,  or  1000  c.c., 
suffered  therefore  a  contraction  of  380  c.c.  (1000  c.c.  — 
620  c.c.).  This  contraction,  due  to  the  removal  of  the 
aqueous  vapor  from  the  side  of  the  products,  corresponds 
to  2  volume-units;  consequently  1  volume-unit  for  this 
reaction  is  equal  to  190  c.c.,  and  the  volume  of  ethylene 
concerned  (1  volume-unit)  is  also  190  c.c.  We  may  as- 
sume, accordingly,  the  presence  of  10  c.c.  of  some  inert  gas 
(e.g.  nitrogen)  in  the  original  volume  (200  c.c.)  of  ethylene 
taken.  This  signifies  that  the  sample  of  ethylene  was 
only  95  per  cent  pure. 

Considerations  into  which  Different  Volume-Units  may 
Enter.  —  When  a  contraction  arises  from  the  combined 
effect  of  two  or  more  reactions  it  may  be  impossible  to 
derive  any  one  of  the  volume-units,  for,  unless  the  rela- 
tive volumes  of  the  several  gases  going  to  produce  the 
contraction  is  known,  we  cannot  properly  apportion  this 
volume  of  contraction  between  the  several  equations. 
In  other  words,  the  volume-unit  may  be  different  for 
every  equation  that  is  brought  into  consideration.  In 
such  cases  as  these  it  is  necessary  to  bring  the  final 
products  under  new  conditions,  whereby  further  conden- 
sations or  absorptions  can  take  place,  and  the  possibility 
of  involving  separately  only  one  product  from  one  of  the 


134  CHEMICAL  CALCULATIONS 

reactions  at  a  time  made  likely.  An  individual  reaction 
thus  concerned  yields  itself  at  once  to  the  determination 
of  its  specific  volume-unit.  By  reference  of  this  volume- 
unit  to  the  several  reactions  of  a  problem  it  is  often  pos- 
sible to  determine  other  volume-units.  Still  further  con- 
densations or  absorptions  through  new  conditions  may  be 
necessary,  however,  to  aid  in  determining  the  volume-units 
of  certain  equations  when  a  large  number  of  substances 
are  present  in  the  original  mixture.  When  all  of  these 
values  are  found  the  solution  of  the  problem  is  compara- 
tively simple. 

Example  62.  —  A  mixture  of  nitrogen,  hydrogen  and  carbon 
monoxide,  450  c.c.  in  volume,  was  exploded  with  an  excess  of 
oxygen,  250  c.c.  After  the  explosion  the  volume  of  gaseous 
product  measured  500  c.c.  With  the  removal  of  the  aqueous 
vapor  the  volume  of  product  measured  400  c.c.  What  was  the 
volume  of  each  component  in  the  original  mixture?  A  tem- 
perature of  100°  and  atmospheric  pressure  considered  constant. 

The  two  equations  are  as  follows: 

2  H2   +  02  =  2  H20. 
Molecular  volumes:    2     +  1    =  2  +  (1vol.  contr.). 

2  CO  +  O2  =  2  CO2. 
Molecular  volumes:    2     +1=2  +  (1  vol.  contr.). 

In  this  problem  the  dry  gaseous  product  measured 
400  c.c.,  i.e.,  by  loss  of  the  aqueous  vapor,  a  contraction  of 
100  c.c.  was  recorded.  This  contraction  is  due  alone  to  the 
reaction  of  hydrogen  with  oxygen,  and  stands  for  the  loss 
of  2  volume-units  of  aqueous  vapor  in  the  first  volume 
equation.  Therefore  1  volume-unit  in  this  equation  is 
equivalent  to  50  c.c.,  and  2  volume-units,  representing 
the  actual  amount  of  hydrogen  concerned  in  the  reaction, 
correspond  to  100  c.c. 


COMBINATIONS   BETWEEN   GASES   BY   VOLUME       135 

In  the  combustion  of  hydrogen  the  contraction  actually 
possible  corresponds  to  1  volume-unit  of  the  reaction. 
This  volume-unit  has  just  been  determined  as  50  c.c. 
The  total  contraction  in  the  problem,  due  to  the  two 
reactions  and  without  elimination  of  aqueous  vapor,  is 
recorded  as  200  c.c.;  hence  the  difference,  200  c.c.  —  50  c.c., 
or  150  c.c.,  represents  the  contraction  due  to  the  reaction 
between  carbon  monoxide  and  oxygen.  This  contraction, 
as  shown  in  the  second  volume  equation  above,  is  equiva- 
lent to  1  volume-unit.  Accordingly  the  volume  of  the 
carbon  monoxide  present  —  2  volume-units  —  must  have 
been  2  X  150  c.c.  or  300  c.c.  All  told,  we  derive  the 
following  composition  for  the  original  mixture:  100  c.c. 
of  hydrogen,  300  c.c.  of  carbon  monoxide  and  50  c.c.  of 
nitrogen  (considered  in  these  problems  as  an  inert  gas). 

The  Algebraic  Method  for  the  Calculation  of  Volume- 
Units.  —  More  complicated  examples  may  be  given,  but 
throughout  all  the  same  principles  hold.  We  must  first 
determine  the  volume-unit  for  each  of  the  reactions  under 
consideration  before  we  can  calculate  the  several  volumes 
concerned  in  the  problem.  When  a  change  in  volume  of 
product  is  found  to  result,  not  from  one  single  reaction, 
but  from  a  combination  of  several,  and  when  also,  through 
new  conditions  presented,  still  further  condensations  occur, 
which  in  turn  are  found  to  be  due  to  a  combination  of 
reactions,  we  find  it  necessary  to  give  the  volume-unit  for 
each  reaction  an  algebraic  symbol,  and  solve  for  the  value 
of  each  unit  from  the  several  equations  that  may  be 
constructed. 

Example  63.  —  A  mixture  of  nitrogen,  carbon  monoxide, 
methane  and  ammonia,  amounting  to  1250  c.c.,  was  exploded 
with  an  excess  of  oxygen  (1100  c.c.),  and  the  volume  of  product 
found  to  measure  2450  c.c.  A  temperature  of  100°  and  760  mm. 
constant  throughout.  After  withdrawal  of  aqueous  vapor  by 


136  CHEMICAL  CALCULATIONS 

absorption,  the  volume  of  the  dry  gases  at  the  recorded  constant 
conditions  was  950  c.c.  After  removal  of  carbon  dioxide  (by 
passing  gases  over  lime)  the  volume  read  600  c.c.  What  was 
the  volume  of  each  component  in  the  original  mixture? 

The  equations  for  the  three  reactions  are  as  follows; 

2  CO  +    O2  =  2  C02 
Molecular  volumes :      2     +1      =  2  +  (1  vol.  contr.)e 

CH4+  2O2=CO2   +  2H2O 
Molecular  volumes:    1        +2=1       +2. 

4NH3+302=2N2  +  6H20 
Molecular  volumes :      4     +3      =  2  +  6  —  (1  vol.  expan. ) . 

An  expansion  follows  the  combustion  of  ammonia 
with  oxygen.  A  contraction  due  to  the  combustion  of 
carbon  monoxide  is  not  sufficient  to  make  up  for  expan- 
sion in  the  former  case,  since  the  final  volume,  2450  c.c., 
is  100  c.c.  larger  than  the  volume  of  the  original  mixture 
of  gases.  As  the  contraction  and  expansion  are  each 
represented  by  a  volume-unit  in  their  respective  reac- 
tions, we  draw  the  conclusion  that  the  volume-unit  in  the 
ammonia  reaction  is  larger  by  100  c.c.  than  the  volume- 
unit  in  the  carbon  monoxide  reaction.  The  withdrawal 
of  aqueous  vapor  is  similarly  distributed  over  two  reac- 
tions, as  is  also  the  withdrawal  of  carbon  dioxide. 

The  data  from  this  problem  do  not  yield  positive 
information  in  regard  to  any  single  volume-unit  of  any 
reaction.  Combinations  between  the  reactions  are  of 
course  easily  intelligible.  Thus  the  withdrawal  of  the 
water  removes  6  volume-units  from  the  ammonia  reac- 
tion and  2  volume-units  from  the  methane  reaction. 
With  all  the  volume-units  unknown  we  may  profitably 
study  the  combination  of  these  units  with  the  idea  of 
deriving  some  one  of  them  and  eventually  all.  For  this 


COMBINATIONS  BETWEEN  GASES  BY  VOLUME   137 

purpose  let  x  represent  the  volume-unit  of  the  carbon 
monoxide  reaction,  y  the  volume-unit  of  the  methane 
reaction,  and  z  the  volume-unit  of  the  ammonia  reac- 
tion. From  the  preceding  remarks  and  a  study  of  the 
reactions  themselves  we  may  now  draw  up  the  following 
equations. 

The  expansion  is  due  to  the  larger  value  of  z  compared 
with  x: 

(a)   z  -  x  =  100. 

The  volume  of  aqueous  vapor  is  represented  by  1500  c.c. 
(2450  c.c.  —  950  c.c.),  and  is  therefore  expressed  by 

(b)    6  z  +  2  y  =  1500. 

The  volume  of  carbon  dioxide  corresponds  to  2  volume- 
units  of  x  and  1  volume-unit  of  y  and  is  represented  by 
350  c.c.  (950  c.c.  -  600  c.c.). 

(c)    y  +  2  x  =  350. 
By  combining  equations  (b)  and  (c)  to  eliminate  y, 

(b)     6  z  +  2  y  =  1500. 

Twice  (c)  or  4  x  +  2  y  =    700,  and  subtracting 

6  z  —  4  x  =    800,  and  subtracting 

four  times  (a)  or  4  z  —  4  x  =    400 
2~T~          =    400 
z  =    200. 

With  the  volume-unit  of  the  ammonia  equation  thus 
derived  and  equal  to  200  c.c.,  we  have  only  to  derive  the 
value  of  x  from  equation  (a)  as  equal  to  100  c.c.  and  then 
in  turn  from  (b)  we  derive  the  value  of  y  as  equal  to  150  c.c. 
Substituting  these  values  in  their  proper  equations  the 
volume  of  carbon  monoxide  (2  x)  is  found  to  be  200  c.c., 
that  of  methane  (y)  150  c.c.,  and  that  of  ammonia  (4  z) 
800  c.c.,  making  in  all  1150  c.c.  Therefore  the  remaining 


138  CHEMICAL   CALCULATIONS 

volume  of  the  original  1250  c.c.,or  100  c.c.,is  calculated  as 
nitrogen. 

Considerations  into  which  the  Formation  of  Non-gase- 
ous Substances  Enter.  —  In  the  study  of  those  molecular 
equations  in  which  a  number  of  non-gaseous  substances 
come  under  consideration  we  have  simply  another  case 
of  removal  of  molecular  quantities. 

Example  64.  —  When  phosphorus  is  burned  in  a  vessel  con- 
taining 100  c.c.  of  nitrous  oxide,  what  will  be  the  volume  of 
nitrogen  left?  The  room  temperature  and  pressure  constant 
throughout. 

The  solid  phosphorus  pentoxide  as  well  as  the  phos- 
phorus fall  out  of  consideration  in  the  volume  equation: 

5  N20  +    P2      =  5  N2  +  P205. 
5  N2O  =5  N2. 

Molecular  volumes:         5        =5. 

There  is,  therefore,  no  change  in  volume  and  100  c.c.  will 
represent  also  the  volume  of  the  nitrogen. 

Again  we  may  consider  the  reaction  of  hydrogen 
chloride  upon  a  carbonate: 

Example  65.  —  An  excess  of  sodium  hydrogen  carbonate  was 
placed  in  a  vessel  containing  100  c.c.  of  hydrogen  chloride. 
Calculate  the  volume  of  dry  carbon  dioxide  liberated  in  the 
reaction.  Room  temperature  and  pressure  a  constant. 

NaHCO3  +  HC1  =  NaCl  +  H2O  +  C02. 
Molecular  volumes :      1     =  1+1. 

We  have  in  this  volume  equation  1  volume  of  hydrogen 
chloride  liberating,  from  1  molecule  of  sodium  hydrogen 
carbonate,  1  volume  of  carbon  dioxide  and  1  volume  of 
aqueous  vapor.  In  the  dry  state,  therefore,  the  volume 


COMBINATIONS   BETWEEN   GASES   BY   VOLUME        139 

relations  of  carbon  dioxide  and  hydrogen  chloride  are  as 
1:1.  In  the  case  of  the  normal  sodium  carbonate  the 
volume  relations  of  these  dry  gases  are  as  1  :  2. 

Na2CO3  +  2  HC1  =  2  Nad  +  H2O  +  C03. 
2      =  1. 

The  proportional  values  of  carbon  dioxide  and  hydrogen 
chloride  are  different  in  the  two  cases.  This,  however,  is 
dependent  upon  the  nature  of  the  substances  concerned 
and  is  easily  explainable  by  a  study  of  the  two  reactions. 
Whatever  be  the  complexity  of  the  examples  here  pre- 
sented, they  nevertheless  may  be  made  to  conform  to 
very  simple  interpretations  when  once  the  molecular 
equations  for  the  reactions  are  constructed  and  a  study 
of  the  volume  relations  undertaken. 

PROBLEMS. 

212.  In  the  combustion  of  400  c.c.  of  hydrogen  with  oxygen, 
what  volume  of  oxygen  will  be  required,  and  what  volume  of 
aqueous  vapor  will  result?    A  temperature  of  100°  and  a  pres- 
sure of  760  mm.  constant  throughout.     Ans.  200  c.c.  oxygen. 

400  c.c.  vapor. 

213.  A  mixture  of  300  c.c.  of  hydrogen  and  200  c.c.  of  oxygen 
was  exploded  by  electric   spark.     What   was  the   volume   of 
product?    A  temperature  of  100°  and  pressure  of  760  mm. 
constant  throughout.  Ans.  350  c.c. 

214.  A  mixture  of  300  c.c.  of  hydrogen  and  130  c.c.  of  oxy- 
gen was  exploded.     What  was  the  volume  of  product?    Tem- 
perature of  100°  and  pressure  of  760  mm.  constant  throughout. 

Ans.  300  c.c. 

215.  A  mixture  of  420  c.c.  of  hydrogen  and   180  c.c.  of 
oxygen  was  exploded.     What  was  the  volume  of  product  after 
the  removal  of  aqueous  vapor  (by  absorption  with  phosphorus 
pentoxide)?    Temperature  of  100°  and  pressure  of  760  mm.  a 
constant.  Ans.  60  c.c. 


140  CHEMICAL  CALCULATIONS 

216.  A  mixture  of  300  c.c.  of  methane  and  150  c.c.  of  oxy- 
gen was  exploded.     What  was  the  volume  of  the  product  after 
the  removal  of  aqueous  vapor  (by  absorption)  ?    Temperature 
of  100°  and  pressure  of  760  mm.  constant  throughout. 

Ans.  300  c.c. 

217.  A  mixture  of  250  c.c.  of  carbon  monoxide  and  120  c.c. 
of  oxygen  was  exploded.     Calculate  the   volume   of  gaseous 
product.    Temperature  and  pressure  a  constant. 

Ans.  250  c.c. 

218.  A  mixture  of  320  c.c.  of  carbon  monoxide  and  180  c.c. 
of  oxygen  was  exploded.     Calculate  the  volume   of  gaseous 
product  after  the  removal  of  the  carbon  dioxide  (absorption  by 
lime).    Temperature  and  pressure  a  constant.        Ans.  20  c.c. 

219.  A  mixture  of  200  c.c.  of  methane  and  300  c.c.  of  carbon 
monoxide  was  exploded  with  600  c.c.  of  oxygen  (excess).     Cal- 
culate the  volume  of  gaseous  product.     Calculate,  also,  the 
volume  of  product  when  deprived  of  aqueous  vapor.    Tem- 
perature of  100°  and  pressure  of  760  mm.  a  constant  throughout. 

Ans.  950  c.c.  (moist). 
550  c.c.  (dry). 

NOTE.  —  Since  the  oxygen  is  in  excess  we  know  the  calculation 
is  to  be  based  upon  the  CH4  and  CO.  The  oxygen  left  unconsumed 
in  these  two  reactions  is  readily  determined  by  difference. 

220.  A  mixture  of  80  c.c.  of  methane  and  200  c.c.  of  oxygen 
was  exploded  over  mercury.     Calculate  the  volume  of  product. 
A  temperature  of  20°  and  barometric  pressure  of  757.4  mm. 
constant  throughout.  Ans.  122.83  c.c. 

NOTE.  — The  volume  of  dry  gaseous  product,  at  20°  and  757.4  mm., 
is  simply  to  be  changed  to  accord  with  the  observed  conditions  in 
the  presence  of  aqueous  vapor.  These  experiments  are  usually  carried 
out  over  mercury.  The  excess  of  water  condenses  of  course  upon 
the  mercury  column.  If  an  appreciable  quantity  were  present  allow- 
ance for  its  pressure  upon  the  column  would  need  to  be  made. 

221.  A  mixture  of  400  c.c.   of  hydrogen  and  300  c.c.   of 
oxygen  was  exploded.     Calculate  the  volume  of  product.     A 
temperature  of  25°  and  pressure  of  753.6  mm.  constant. 

Ans.  103.23  c.c. 


COMBINATIONS    BETWEEN   GASES   BY    VOLUME        141 

222.  A  mixture  of  200  c.c.  of  carbon  monoxide  and  300  c.c. 
of  oxygen  was  exploded  over  mercury.     Calculate  the  volume 
of  product.    A  temperature  of  27°  and  pressure  of  740  mm. 
constant  throughout.  Ans.  400  c.c. 

223.  A  mixture  of  oxygen  and  hydrogen,  measuring  150  c.c. 
in  volume,  was  exploded  by  means  of  an  electric  spark.     The 
volume  of  product  measured   125  c.c.     Temperature  of  100° 
and  atmospheric  pressure  a  constant.     Calculate  the  volumes 
of  hydrogen  and  oxygen  concerned  in  the  reaction.     The  resid- 
ual gas  is  of  course  the  excess  of  either  the  hydrogen  or  oxygen 
present.  Ans.  50  c.c.  hydrogen. 

25  c.c.  oxygen. 

224:.  A  mixture  of  oxygen  and  hydrogen,  measuring  150  c.c. 
in  volume,  was  exploded.  The  volume  of  product  measured 
76.47  c.c.  Temperature  of  17°  and  pressure  of  754.4  mm.  a 
constant.  Calculate  the  volumes  of  hydrogen  and  oxygen  con- 
cerned in  the  reaction.  Ans.  50  c.c.  hydrogen. 

25  c.c.  oxygen. 

335.  250  c.c.  of  dry  air  were  mixed  with  150  c.c.  of  hydrogen 
(an  excess)  and  exploded.     The  volume  of  product  was  350  c.c. 
Calculate  the  percentage  of  oxygen  (by  volume)  in  the  sample 
of  air.    Temperature  of  100°  and  atmospheric  pressure  a  con- 
stant. Ans.  20  per  cent. 

336.  300  c.c.  of  dry  air  were  mixed  with  250  c.c.  of  hydrogen 
(an  excess)   and  exploded.    The  volume  of  the  dry  gaseous 
product  was  361  c.c.     Calculate  the  percentage  of  oxygen  (by 
volume)  in  the  sample  of  air.    Temperature  of  20°  and  pressure 
of  745  mm.  a  constant.  Ans.  21  per  cent. 

337.  A  mixture  of  hydrogen  sulphide  with  an  excess  of 
oxygen  measured  350  c.c.  at  100°  and  750  mm.  pressure.     After 
explosion  (with  complete  combustion)  the  volume  of  dry  gase- 
ous product  read  260  c.c.  at  these  same  conditions.     Calculate 
the  volume  of  hydrogen  sulphide  in  the  mixture.   Ans.  60  c.c. 

338.  A  mixture  of  hydrogen  sulphide  with  an  excess  of 
oxygen  measured  200  c.c.   at  20°  and   747.4  mm.   pressure. 
After  explosion   (with  complete   combustion)    the   volume   of 
product  read  143.3  c.c.  at  these  same  conditions.     Calculate 
the  volume  of  hydrogen  sulphide  in  the  mixture.   Ans.  40  c.c. 


142  CHEMICAL   CALCULATIONS 

229.  A  mixture  of  acetylene,  C2H2,  with  an  excess  of  oxygen 
measured  350  c.c.  at  25°  and  745  mm.  pressure.     After  explo- 
sion the  volume  of  dry  gaseous  product  read  275  c.c.  at  the 
same  conditions.     Calculate  the  volume  of  acetylene  in  the 
mixture.  Ans.  50  c.c. 

230.  A  mixture  of  acetylene,  C2H2,  with  an  excess  of  oxygen 
measured  240  c.c.  at  24°  and  752.4  mm.  pressure.     After  ex- 
plosion the  volume  of  product  read  221.8  c.c.  at  28°  and  750.1 
mm.  pressure.     Calculate  the  volume  of  acetylene  in  the  mix- 
ture. Ans.  20  c.c. 

331.  An  excess  of  oxygen,  400  c.c.,  was  admitted  into  a 
volume  of  ammonia  (containing  an  impurity  of  air,  i.e.,  nitro- 
gen and  oxygen)  which  measured  240  c.c.,  and  the  mixture,  at 
22°  and  745.2  mm.  pressure,  exploded.  The  volume  of  product 
measured  415.1  c.c.  at  24°  and  the  same  pressure.  Calculate 
the  percentage  purity  of  the  sample  of  ammonia. 

Ans.  80  per  cent. 

232.  Into  a  stoppered  tube  (90  c.c.  in  capacity)  filled  with 
chlorine,   a   small   quantity   of   concentrated   ammonia   water 
(10  c.c.  or  an  excess)  was  admitted.     After  shaking,  the  mouth 
of  the  tube  was  opened  under  a  dilute  acid  solution  contained 
in  a  tall  cylinder.     The  excess  of  ammonia,  together  with  the 
hydrogen  chloride,  formed  in  the  reaction,  were  thus  removed. 
The  residual  volume  of  gas  may  be  considered  as  nitrogen. 
Calculate  the  volume  of  nitrogen  in  the  tube.     The  room  tem- 
perature and  pressure  considered  a  constant  throughout. 

Ans.  30  c.c. 

233.  A  mixture  of  carbon  monoxide  and  methane  contain- 
ing also  nitrogen  measured  260  c.c.     Into  this  mixture  was 
admitted  an  excess  of  oxygen,  300  c.c.,  and  the  whole  exploded. 
After  the  explosion  the  volume  of  product  measured  520  c.c., 
but  after  the  removal  of  aqueous  vapor  (by  absorption)  from  the 
gaseous  product  the  volume  measured  320  c.c.     Calculate  the 
volume  of  each  component  in  the  original  mixture.     A  tem- 
perature of  100°  and  atmospheric  pressure  constant. 

Ans.    80  c.c.  CO. 
100  c.c.  CH4. 
80  c.c.  N. 


COMBINATIONS   BETWEEN   GASES   BY    VOLUME        143 

234.  A  sample  of  water  gas  containing  an  impurity  of  air 
measured   240   c.c.   in   volume.     Into  this  was  admitted   an 
excess  of  oxygen,  400  c.c.,  and  the  mixture  exploded.    The 
volume  of  product  measured  540  c.c.     After  removal  of  aque- 
ous vapor   (by  absorption)   the  volume  of  product  measured 
440   c.c.     Calculate   the   percentage  of  hydrogen  and   carbon 
monoxide  in  the  sample.     A  temperature  of  100°  and  atmos- 
pheric pressure  constant  throughout. 

Ans.  41.67  per  cent  H. 
41.67  per  cent  CO. 

235.  500  c.c.  of  a  mixture  of  ammonia,  methane  and  nitrogen 
were  mixed  with  500  c.c.  of  oxygen  and  the  mixture  exploded. 
The  volume  of  the  product  measured  1050  c.c.     A  constant 
temperature    of    100°   and   pressure    of   760   mm.    maintained 
throughout.     Upon  cooling  the  gaseous  product  to  remove  the 
aqueous  vapor  and  recalculation  of  the  volume  of  dry  gas  to 
the  original  conditions  (100°),  the  volume  measured  550  c.c. 
What  was  the  composition  of  the  original  mixture? 

Ans.  100  c.c.  CH4. 
200  c.c.  NH8. 
200  c.c.  N. 

236.  A   mixture   of  methane,   hydrogen   sulphide   and   air 
measured  300  c.c.     When  mixed  with  an  excess  of  oxygen, 
500  c.c.,  and  exploded  the  volume  of  product  measured  720 
c.c.     A  temperature  of  100°  and  atmospheric  pressure  constant 
throughout.     Upon  cooling  the  gaseous  product  to  the  room 
temperature  and  recalculation  of  the  volume  of  dry  gas  to  the 
original  conditions,  the  volume  measured  320  c.c.    Calculate  the 
composition  of  the  original  mixture.  Ans.  120  c.c.  CH4. 

160  c.c.  H2S. 
20  c.c.  air. 

237.  A  mixture  of  methane  and  acetylene  measuring  500 
c.c.  was  mixed  with  1500  c.c.  of  oxygen  (an  excess)  and  ex- 
ploded.    The  volume  of  product  measured  1850  c.c.     A  tem- 
perature  of   100°  and  atmospheric  pressure   constant.     Upon 
removal   of  aqueous   vapor   (by   condensation  and   recalcula- 
tion to  the  original  conditions)  the  volume  measured  1150  c.c. 
Calculate  the  composition  of  the  original  mixture. 

Ans.  200  c.c.  CH4. 
300  c.c.  C3H2. 


144  CHEMICAL   CALCULATIONS 

238.  A  mixture  of  hydrogen,  methane  and  nitrogen  measur- 
ing 350  c.c.  was  exploded  with  an  excess  of  oxygen,  500  c.c. 
After  the  explosion  the  volume  of  dry  gaseous  product  meas- 
ured 475  c.c.     Upon  removal   of   carbon  dioxide   the  volume 
of  gas  measured  400  c.c.     Calculate  the  composition  of  the 
original  mixture.     Room  temperature  and  pressure  a  constant. 

Ans.  150  c.c.  H. 

75  c.c.  CH4. 
125  c.c.  N. 

239.  A  mixture  of  cyanogen,  methane  and  nitrogen  meas- 
uring 200  c.c.  in  volume  was  exploded  with  an  excess  of  oxygen, 
500  c.c.      After  the  explosion  the  volume  of  product  meas- 
ured  619.6  c.c.     Upon  removal  of  carbon  dioxide   (by  lime) 
the  volume  of  dry  gas  measured  430  c.c.     Calculate  the  com- 
position of  the  original  mixture.     A  temperature  of  25°  and 
pressure  of  746.6  mm.  constant  throughout. 

Ans.  50  c.c.  CH4. 
60  c.c.  C2N3. 
90  c.c.  N. 

240.  A  liter  of  methane  contaminated  with  the  vapor  of 
carbon  disulphide  was  mixed  with  an  excess  of  oxygen,  3000  c.c., 
and   exploded.     The   volume   of  product   measured   3840   c.c. 
Calculate  the  percentage  impurity  of  the  methane.     A  tempera- 
ture of  100°  and  atmospheric  pressure  constant  throughout. 

Ans.  16  per  cent. 

241.  An  excess  of  oxygen  was  admitted  to  a  vessel  contain- 
ing hydrogen  and  ammonia  and  the  mixture  exploded.     After 
the  explosion  there  was  observed   no   change   in   volume    of 
product  from  that  of  the  mixture.     Calculate  the  percentage 
composition  of  the  hydrogen  and  ammonia  mixture.     A  tem- 
perature of  100°  and  atmospheric  pressure  constant. 

Ans.  33£  per  cent  H. 
66|  per  cent  NH3. 

242.  A  mixture   of  ethylene  and  ammonia,   contaminated 
with  air  measured  600  c.c.     When  this  volume  of  gas  was  mixed 
with  an  excess  of  oxygen,  2000  c.c.,  and  exploded,  the  volume 
of    product    measured    2640   c.c.     Upon   removal   of   aqueous 
vapor  (by  cooling  and  recalculation)  the  volume  of  dry  gas 


COMBINATIONS   BETWEEN   GASES   BY   VOLUME        145 

measured  2120  c.c.  Calculate  the  composition  of  the  original 
mixture.  A  temperature  of  100°  and  atmospheric  pressure 
constant.  Am.  160  c.c.  NH3. 

140  c.c.  C2H4. 

300  c.c.  air. 

243.  A  mixture    of   ethylene  and  ammonia   contaminated 
with  air  measured  400  c.c.      When  this  volume  of  gas  was 
mixed  with  an  excess  of  oxygen,  1200  c.c.,  and  exploded,  the 
volume  of   product  measured    1293.6   c.c.     Upon   removal   of 
carbon  dioxide    (by  lime)   the   volume   of  dry  gas  measured 
1050  c.c.     Calculate  the  composition  of  the  original  mixture. 
A  temperature   of  26°  and  pressure   of  745.6  mm.   constant 
throughout.  Ans.  100  c.c.  C2H4. 

120  c.c.  NH3. 
180  c.c.  air. 

244.  Into  a  mixture  of  cyanogen,  ethylene  and  nitrogen  was 
admitted  a  large  excess  of  oxygen  and  the  mixture,   which 
measured  1600  c.c.  at  22°  and  753  mm.  pressure,  exploded. 
After  the  explosion  the  volume  of  product  measured  1257  c.c. 
at  24°  and  746  mm.  pressure.     Upon  removal  of  carbon  dioxide 
and  moisture  the  volume  of  dry  gas  (still  at  24°  and  746  mm.) 
measured   609.7  c.c.     Calculate  the  amount  of  cyanogen  and 
ethylene  in  the  original  mixture.  Ans.  100  c.c.  C2N2. 

200  c.c.  C2H4. 

345.  A  mixture  of  hydrogen  and  ammonia  contaminated 
with  air  measured  400  c.c.  Into  this  was  admitted  an  excess 
of  oxygen,  600  c.c.,  and  the  final  mixture  exploded.  After  the 
explosion  the  volume  of  product  measured  1025  c.c.  Upon 
removal  of  aqueous  vapor  (by  absorption)  the  volume  read 
675  c.c.  Calculate  the  composition  of  the  original  mixture.  A 
temperature  of  100°  and  atmospheric  pressure  constant  through- 
out. (Cf.  Ex.  63.)  Ans.  50  c.c.  H. 

200  c.c.  NH8. 
150  c.c.  air. 

246.  A  mixture  of  hydrogen,  methane,  carbon  monoxide 
and  nitrogen  measured  500  c.c.  Into  this  was  admitted  an 
excess  of  oxygen,  900  c.c.,  and  the  final  mixture  exploded. 
After  the  explosion  the  volume  of  product  measured  1250  c.c. 
Upon  removal  of  aqueous  vapor  (by  absorption)  the  volume 
measured  1000  c.c.  Upon  removal  of  carbon  dioxide  (by 


146  CHEMICAL   CALCULATIONS 

lime)  the  volume  measured  725  c.c.  Calculate  the  composition 
of  the  original  mixture.  A  temperature  of  100°  and  atmos- 
pheric pressure  constant.  (Cf.  Ex.  63.)  Ans.  100  c.c.  H. 

200  c.c.  CO. 
75  c.c.  CH4. 

125  c.c.  N. 

347.  A  mixture  of  carbon  monoxide  and  methane  contami- 
nated with  air  and  measuring  400  c.c.  was  exploded  with  an 
excess  of  oxygen,  1000  c.c.  The  volume  of  product  measured 
1025.4  c.c.  Upon  removal  of  the  carbon  dioxide  (by  lime)  the 
volume  of  dry  gas  measured  725  c.c.  Calculate  the  composi- 
tion of  the  original  mixture.  A  temperature  of  21°  and  pres- 
sure of  745.5  mm.  constant  throughout.  Ans.  100  c.c.  CO. 
(Cf.  Ex.  63.)  175  c.c.  CH4. 

125  c.c.  air. 

248.  A  mixture  of  cyanogen,  methane  and  hydrogen  con- 
taminated with  air  measured  800  c.c.  An  excess  of  oxygen, 
2000  c.c.,  was  admitted  into  the  vessel  containing  this  mixture 
and  the  final  mixture  exploded.  After  the  explosion  the  vol- 
ume of  product  measured  2750  c.c.  After  removal  of  aqueous 
vapor  (by  cooling  and  recalculation)  the  volume  measured  2250 
c.c.  After  removal  of  carbon  dioxide  (by  lime)  the  final  vol- 
ume measured  1450  c.c.  Calculate  the  composition  of  the 
original  mixture.  A  temperature  of  100°  and  atmospheric 
pressure  constant.  Ans.  100  c.c.  H. 

(Cf.  Ex.  63.)  200  c.c.  CH4. 

300  c.c.  C2N2. 

100  c.c.  air. 

349.  An  excess  of  oxygen,  800  c.c.,  was  added  to  a  mixture 
of  ammonia,  ethylene  and  hydrogen,  310  c.c.  in  volume,  and 
the  final  mixture  exploded.  After  the  explosion  the  volume 
of  product  measured  648.35  c.c.  Upon  removal  of  carbon 
dioxide  (by  lime)  the  volume  of  dry  gas  measured  435  c.c. 
Calculate  the  composition  of  the  original  mixture.  A  tempera- 
ture of  18°  and  pressure  of  746  mm.  constant  throughout. 
(Cf.  Ex.  63.)  Ans.  50  c.c.  H. 

100  c.c.  C2H4. 

160  c.c.  NH3. 

NOTE.  —  The  contraction  due  to  the  reactions  is  here  masked 
in  the  contraction  from  the  loss  of  aqueous  vapor.  Construct  the 


COMBINATIONS   BETWEEN    GASES   BY   VOLUME        147 

third  equation  from  the  volume  of  oxygen  required  in  the  several 
combustions,  bearing  in  mind  that  the  volume,  365  c.c.  (800  c.c.  — 
435  c.c.),  which  nominally  represents  this  quantity  in  the  problem, 
is  in  fact  less  than  the  actual  volume  by  that  volume  of  nitrogen  left 
free. 

250.  A  mixture  of  methane  and  ethylene  measuring  300  c.c. 
was  exploded  with  an  excess  of  oxygen,  1500  c.c.     After  the 
explosion  the  volume  of  dry  gaseous  product   (calculated  to 
observed   conditions  of  temperature   and   pressure)    measured 
1200  c.c.     In  order  to  determine  the  volume  of  oxygen  actually 
consumed  in  this  combustion  and  thus  present  data  upon  which 
a  second  equation  may  be  constructed,  an  excess  of  hydrogen, 
2600  c.c.,  was  admitted  to  the  vessel  containing  the  dry  gaseous 
product.     After  an  explosion  the  volume  of  this  second  product 
measured    1400   c.c.    (calculated   as   above,   with   absence   of 
aqueous  vapor).     Calculate  the  amount  of  methane  and  ethyl- 
ene in  the  original  mixture.     The  room  temperature  and  pres- 
sure remained  constant  throughout.  Ans.  200  c.c.  CH4. 
(Cf.  Ex.  63.)  100  c.c.  C2H4. 

251.  A  volume  of  nitric  oxide,  NO,  measuring  400  c.c.  was 
required  for  the  combustion  of  a  definite  weight  of  phosphorus. 
What  volume  of  nitrogen  remained  free?    The  room  tempera- 
ture and  pressure  a  constant.  Ans.  200  c.c. 

252.  A  quantity  of    phosphorus  was   burned   in  a   vessel 
over  water  holding  222.7  c.c.   of   nitrous   oxide,    N20,   meas- 
ured at  20°  and  750  mm.  pressure.     What  volume  of  nitrogen 
remained,  the  temperature  and  pressure  constant? 

Ans.  222.7  c.c. 

253.  What  volume   of  water  gas  is  theoretically  possible 
from  the  action  of  1  liter  of  steam  upon  heated  coke?    Tempera- 
ture and  pressure  a  constant.  Ans.  2000  c.c. 

254.  In  the  decomposition  of  methane  by  chlorine  what 
volume  of  hydrogen  chloride  corresponds  to  1  volume  of  meth- 
ane?   Temperature  and  pressure  constant.      Ans.  4  volumes. 

255.  A  mixture  of  400  c.c.  of  methane  and  1000  c.c.  of  chlo- 
rine was  exploded.     Calculate  the  volume  of  gaseous  product, 
temperature  and  pressure  a  constant.  Ans.  1800  c.c. 

256.  What  volume  of  gaseous  product  may  be  obtained  in 
the  decomposition  of  100  c.c.  of  ammonia  by  heated  cupric 
oxide?    Temperature  and  pressure  a  constant. 

Ans.  200  c.c. 

(Steam  +  N). 


CHAPTER  XII. 
COMPLEX   EQUATIONS. 

THE  study  of  chemical  reactions  as  outlined  in  Chapter 
IX  was  not  inclusive  of  those  examples  where  a  change 
in  the  valence  of  an  element  occurs.  Although  these 
changes  may  complicate  the  matter  of  drawing  up  the 
equations,  the  same  principles  as  already  discussed  will 
be  found  to  apply.  The  definite  quantity  of  each  element 
concerned,  whether  or  not  it  exerts  the  same  combin- 
ing capacity  for  other  elements  on  the  two  sides  of  an 
equation,  is  nevertheless  a  constant  for  the  particular 
equation  in  question.  The  exact  application  of  these 
principles  will  rest,  of  course,  upon  the  construction  of  the 
equations  in  their  molecular  form. 

Oxidation  and  Reduction.  —  The  increase  in  valence  or 
combining  capacity*  which  accompanies  an  element  when, 
for  instance,  it  passes  from  a  lower  to  a  higher  oxide,  has 
been  taken  as  the  measure  of  the  degree  of  oxidation. 
The  position  of  oxygen  in  this  connection  might  be 
filled  by  almost  any  non-metallic  element  and  still  we 
may  have  the  means  of  measuring  the  increase  in  com- 
bining capacity  of  a  metallic  element  or  radical  acting  as 
such  toward  a  non-metallic  element,  i.e.,  oxidation.  The 
unit  of  measure,  rather  than  oxygen  with  its  2  com- 
bining capacities,  is  referred  to  an  element  with  but  1 

*  The  author  is  indebted  to  his  colleague,  Professor  S.  Laurence 
Bigelow,  for  this  interpretation  of  valence  as  the  capacity  factor 
of  chemical  energy,  i.e.,  the  combining  capacity  of  a  unit  quantity 
of  an  element  or  radical. 

148 


COMPLEX   EQUATIONS  149 

capacity,  e.g.  hydrogen.  Thus  1  molecule  of  chromic 
oxide,  Cr2IIIO311,  undergoes  oxidation  to  2  molecules  of 
chromium  trioxide,  2  CrVI03",  and  the  oxygen  intake  for 
this  definite  quantity  of  chromium  is  3  oxygen  atoms, 
i.e.,  the  equivalent  of  6  hydrogen  atoms  in  combining 
capacity,  or  6  combining  capacities.  This  amount  for 
2  atoms  of  chromium  is  equivalent  to  3  combining  capac- 
ities for  1  atom  of  chromium  and  thus  this  increase  will 
be  denoted  in  the  second  compound  CrVIO3H. 

The  reverse  process,  or  a  decrease  in  this  combining 
capacity  toward  a  non-metallic  element,  is  known  as 
reduction.  For  example,  ferric  iron  as  Fe111^1  is  reduced 
to  ferrous  iron,  Fe"^1. 

Analogous  to  this  the  increase  in  combining  capacity  of 
a  non-metallic  element,  or  a  radical  acting  as  such,  for 
oxygen  or  some  other  non-metal  will  be  a  measure  of  the 
oxidation,  e.g.  Cl^O11  ->  Cl2VIIO7n.  When,  however,  this 
capacity  of  a  non-metal  is  considered  in  relation  to  its 
combination  with  metallic  groups  the  increase  in  its  com- 
bining capacity  is  in  fact  a  measure  of  its  reduction. 
Thus  potassium  ferricyanide,  K3irFeIII(CN)6I]m,  is  reduced 
to  potassium  ferrocyanide,  K4I[Fe"(CN)6I]IV,  when  the  ferri- 
cyanide radical  increases  its  combining  capacity  toward 
the  metallic  element  potassium.  This  result  is  of  course 
due  to  the  reduction  of  the  ferric  iron  in  the  former 
radical  to  ferrous  iron  in  the  latter.  In  general  the 
increase  in  the  combining  capacity  of  a  non-metallic 
radical  for  a  metallic  one  signifies  a  demand  for  a  lower 
number  of  these  capacities  required  of  an  intra-radical 
metal  for  complete  combination,  or  what  was  previously 
interpreted  as  a  reduction. 

Reaction-Quantities  Involved  in  Oxidation  and  Reduc- 
tion. —  The  change  of  an  element  from  the  non-metallic 
to  the  metallic  groupings,  or  vice  versa,  may  be  accom- 


150  CHEMICAL   CALCULATIONS 

panied  by  a  loss  or  gain  in  its  combining  capacity,  and 
consequently  we  speak  of  it  as  undergoing  a  reduction  or 
oxidation.  Thus  potassium  dichromate,  K2Cr207,  con- 
tains the  chromium  associated  directly  with  oxygen  in 
the  form  of  the  acid  H2Cr04  (or  oxide  OVI03"),  but  by  the 
action  of  a  strong  acid  this  chromium,  with  a  combining 
capacity  of  6,  is  completely  transformed  into  a  chromic 
compound  (where  chromium  is  the  basic  element)  and  then 
has  a  combining  capacity  of  only  3,  Cr2in03".  The  action 
therefore  is  described  as  one  of  reduction. 

In  illustration  of  these  points  a  few  of  the  more  com- 
mon examples  will  be  given.  First  we  may  mention  the 
action  of  hydrogen  sulphide  upon  a  solution  of  ferric 
chloride : 

2  FeCl3  +  H2S  =  2  FeCl2  +  2  HC1  +  S. 
Expressed  by  ions: 

2  Fe"  +  6  Cl'  +  2  H"    +  S" 

=  2  Fe"  +  4  Cl7  +  2  H*  +  2  Cl7  +  S, 

where  the  combining  capacities  are  represented  in  solu- 
tion as  ionic  charges. 

From  the  reaction-quantities  noted  we  may  draw  up 
the  several  ratios  of  proportionality  here  existent.  For 
example  the  amount  of  ferrous  chloride  resulting  from  a 
definite  weight  of  ferric  chloride  is  expressed  by  the 
ratio  Fed 2 /Fed 3,  and  then  again  the  weight  (as  well  as 
the  volume  eventually)  of  hydrogen  sulphide  necessary 
for  this  reduction  of  the  ferric  chloride  is  in  accord  with 
the  ratio  H2S/2  FeCl3.  The  hydrogen  sulphide  molecule 
is  equivalent  therefore  in  its  reducing  action  to  2  units 
of  hydrogen,  whereas  2  molecules  of  ferric  chloride  are 
equivalent  in  oxidizing  action  to  1  oxygen  atom. 

Oxidizing  Action  of  Potassium  Dichromate.  —  As  has 
been  stated,  the  oxidizing  action  of  potassium  dichro- 


COMPLEX   EQUATIONS  151 

mate,  K2O207,  in  the  presence  of  a  strong  acid  and  sub- 
stance capable  of  oxidation,  is  due  to  the  transformation 
of  the  dichromate  (Crvl)  into  a  chromic  compound  (Crin). 
The  oxidizing  power  of  this  substance  is  measured  by 
the  liberation  of  3  atoms  of  oxygen  per  molecule  of 
dichromate,  as  may  be  shown  by  the  following  hypotheti- 
cal equation  for  the  action  with  sulphuric  acid: 

K2Cr207  +  4  H2S04  =  K2SO4  +  Cr2(S04)3  +  4  H2O  +  3  0. 

When  an  active  reducing  agent,  e.g.  ammonium  sul- 
phide, is  present  the  reduction  of  the  dichromate  may  be 
accomplished  without  the  presence  of  an  acid: 

K2Cr207  +  3  (NH4)2S  +  7  H2O 

=  2  Cr(OH)3  +  2  KOH  +  6  NH4OH  +  3  S. 

In  other  cases  some  oxidizable  substance  must  be  added 
to  the  acid  mixture  in  order  that  the  oxygen  capable  of 
liberation  may  be  actually  consumed  and  the  reaction 
proceed. 

A  number  of  substances,  such  as  hydrogen  sulphide, 
sulphur  dioxide,  carbon  monoxide  and  alcohol,  are  readily 
oxidized  in  this  manner  to  sulphur,  sulphur  trioxide, 
carbon  dioxide  and  aldehyde,  respectively.  Thus  with 
hydrogen  sulphide: 

K2Cr207  +  4  H2SO4  +  3  H2S 

=  K2S04  +  Cr2(S04)3+  4  H20  +  3  H20  +38. 

The  presence  of  3  molecules  of  hydrogen  sulphide  is  neces- 
sary for  interaction  with  the  3  atoms  of  oxygen  liberated 
per  molecule  of  dichromate;  the  products  by  this  oxidation 
are  indicated  in  the  equation.  1  molecular  weight  of 
potassium  dichromate  is  here  directly  proportional  to 
3  molecular  weights  of  hydrogen  sulphide,  hence  in  this 


152  CHEMICAL   CALCULATIONS 

reaction  all  values  for- these  substances  must  accord  with 
the  ratio 

3H2S  3  (34.09) 

K2Cr2(y  (        294.2 

From  the  relation  between  the  molecular  weight  of  a 
substance  and  its  volume  as  a  vapor  we  are  able  to  calcu- 
late, for  example,  the  volume  of  hydrogen  sulphide  that 
can  be  oxidized  by  a  corresponding  proportional  weight  of 
the  dichromate. 

Example  66.  —  What  volume  of  hydrogen  sulphide,  at  stand- 
ard conditions,  can  be  oxidized  by  7.36  grams  of  potassium 
dichromate  in  an  acid  solution? 

The  volume  of  hydrogen  sulphide  oxidized  by  1  gram- 
molecular  weight  of  potassium  dichromate  (294.2  grams) 
is  that  volume  corresponding  to  3  gram-molecular  weights 
of  the  gas,  or  67,200  c.c.  (i.e.,  3  X  22,400  c.c.)  at  standard 
conditions.  The  direct  proportionality  between  the  gram- 
molecular  weight  and  the  actual  weight  of  dichromate 
here  given,  determines  also  the  proportionality  between 
the  respective  volumes  of  hydrogen  sulphide  correspond- 
ing to  these  values: 

294.2  :  7.36  =  67,200  :  x, 

in  which  x  represents  the  volume  (1681  c.c.)  of  gas  oxi- 
dized. 

The  weight  of  sulphur  precipitated,  as  well  as  any  other 
substance  present,  may  be  calculated  from  the  standpoint 
of  these  proportional  values  expressed  in  the  equation. 
For  example,  the  relation  between  the  sulphur  and  potas- 
sium dichromate  is  expressed  by  the  ratio : 

3  S  96.21 

K2Cr2O7'  294.2* 


COMPLEX   EQUATIONS  153 

When  hydrochloric  acid  is  used  the  excess  of  acid  is 
itself  oxidized  by  the  liberated  oxygen  with  the  produc- 
tion of  free  chlorine  and  water: 

K2Cr207  +  14  HC1  =  2  KC1  +  2  CrCl3  +  7  H20  +  3  C12. 

Standard  Oxidizing  Solutions  with  Potassium  Bichro- 
mate.—  From  reactions  of  oxidation  with  a  potassium 
dichromate  solution  we  are  able  to  draw  up  the  value  of 
such  a  solution  in  terms  of  the  hydrogen  equivalent. 
1  atom  of  oxygen  will  oxidize  and  combine  with  2  atoms 
of  hydrogen,  hence  the  equivalence  of  3  atoms  of  oxygen 
in  terms  of  hydrogen  is  necessarily  6.  Since  1.01  grams 
of  hydrogen  or  this  unit  equivalent  are  required  per  liter 
of  a  normal  solution,  we  shall  require  here  in  a  normal  solu- 
tion of  potassium  dichromate — that  solution  which  per  liter 
can  furnish  just  sufficient  oxygen  to  unite  with  1.01  grams 
of  hydrogen  —  exactly  one-sixth  of  the  gram-molecular 
weight  of  this  salt,  i.e.,  294.2/6  or  49.03  grams.  In  the 
work  of  titration  with  standard  solutions  of  this  salt  the 
change  in  color  from  red  to  green  indicates  complete 
reduction,  but  for  accurate  determination  of  this  point 
it  is  customary  to  involve  small  portions  of  the  solution 
in  further  reactions. 

Example  67.  —  What  is  the  normality  of  that  potassium 
dichromate  solution,  40  c.c.  of  which  oxidized  0.590  gram  of 
ferrous  sulphate  to  the  ferric  state? 

The  more  commonly  used  ferrous  ammonium  sulphate, 
FeS04(NH4)2SO4 .  6  H2O,  is  here  replaced  by  ferrous 
sulphate,  FeS04,  for  the  sake  of  simplicity.  The  reaction 
for  the  oxidation  of  this  latter  to  ferric  sulphate  involves 
1  oxygen  atom,  with  1  molecule  of  sulphuric  acid,  per  2 
molecules  of  the  ferrous  salt: 

2  FeSO4  +  H2S04  +  O  =  Fe2(SO4)3  +  H2O. 


154  CHEMICAL  CALCULATIONS 

Consequently  to  include  this  action  in  the  equation  which 
represents  the  oxidation  by  means  of  1  molecule  of  potas- 
sium dichromate,  with  its  3  possible  oxygen  atoms,  we 
shall  need  to  multiply  the  equation  above  by  3  and  thus 
obtain  3  O  as  a  reaction-quantity  common  to  both  equa- 
tions : 

K2Cr2O7  +  4  H2SO4  =  K2SO4  +  Cr2(S04)3  +  4  H2O  +  3  0 
6  FeSO4  +  3  H2SO4  +  3  O  =  3  Fe2(SO4)3  +  3  H20. 

It  comes  to  the  same  end  if  we  simply  combine  these 
equations  and  let  this  common  quantity  (3  O)  drop  out, 
when  we  shall  have  the  final  expression: 

K2O2O7  +  6  FeSO4  +  7  H2SO4  = 

K2S04  +  Cr2(S04)3  +  3  Fe2(S04)3  +  7  H20. 

In  either  case  the  reaction-quantities  6  FeS04  and  K2O2O7 
are  directly  proportional  to  each  other,  and  the  ratio 
6  FeSO4/K2Cr2O7,  or  911.52/294.2,  determines  the  ratio 
between  all  weights  which  severally  may  represent  these 
quantities. 

In  the  example  0.590  gram  of  FeSO4  was  oxidized. 
Therefore  the  following  equivalence  in  the  ratios: 

911.52  _  0.590 
294.2  =       x 

The  value  for  x  is  found  to  be  0.1904  gram.  This  weight, 
however,  was  present  in  40  c.c.  of  solvent,  consequently 
in  1  liter  we  should  have  25  X  0.1904  or  4.76  grams  of 
potassium  dichromate.  A  normal  solution  of  this  salt 
contains  49.03  grams  per  liter  (p.  153),  hence  we  have 
a  solution  of  4.76/49.03  normality,  i.e.,  0.0971  normal,  a 
trifle  less  than  N/ 10. 

Oxidizing  Action  of  Potassium  Permanganate.  —  As  a 
second  illustration  the  oxidizing  action  of  potassium  per- 


COMPLEX   EQUATIONS  155 

manganate  is  of  much  importance.  The  action  in  an 
acid  solution  is  analogous  to  that  of  potassium  dichromate, 
only  the  reduction  of  Mnvn  to  Mn11  is  accompanied  here 
with  the  liberation  of  5  atoms  of  oxygen  for  every  2  mole- 
cules of  permanganate: 
2  KMnO4  +  3  H2S04  =  K2S04  +  2  MnSO4  +  3  H2O  +  5  O. 

The  presence  of  some  substance  capable  of  oxidation  is 
necessary  in  order  to  bring  about  this  reaction  in  dilute 
solutions.     We  may  take  again  the  example  of  hydrogen 
sulphide  as  affected  by  an  oxidizing  agent: 
2  KMnO,  +  3  H2SO4  +  5  H2S 

=  K2S04  +  2  MnS04+  3  H2O  +  5  H20  +  5  S. 

From  this  equation  the  relative  amounts  of  hydrogen  sul- 
phide and  potassium  permanganate  involved  will  always 
accord  with  the  ratio  between  their  respective  reaction- 
quantities  : 

5H2S  5  (34.09) 

2KMn04'        2(158.03)' 

When  expressed  by  molecular  equations  the  substances 
here  concerned  may  be  studied  also  in  regard  to  their 
volume  relations. 

Example  68.  —  What  weight  of  potassium  permanganate  will 
be  reduced  by  2800  c.c.  of  hydrogen  sulphide,  at  standard  condi- 
tions of  temperature  and  pressure? 

The  reaction  already  discussed  supplies  the  necessary 
data:  2  gram-molecular  weights  (316.06  grams)  of  the 
permanganate  are  reduced  by  5  X  22,400  c.c.  of  hydrogen 
sulphide.  By  simple  proportion  the  weight  of  perman- 
ganate that  can  be  reduced  by  2800  c.c.  is  derived  as 
follows : 

(5  X  22.400)  :  2800  =  316.06  :  x, 
where  x,  or  7.9  grams,  is  this  weight  of  permanganate. 


156  CHEMICAL   CALCULATIONS 

The  reverse  operation,  or  that  of  determining  the  weight 
(or  volume)  of  hydrogen  sulphide  oxidized  by  a  known 
weight  of  potassium  permanganate,  follows  the  same 
principle. 

Standard  Oxidizing  Solutions  with  Potassium  Perman- 
ganate. —  The  liberation  of  5  atoms  of  oxygen  per  2 
molecules  of  permanganate  also  determines  the  amount 
of  this  salt  that  must  be  present  in  one  liter  of  its  normal 
solution  for  use  in  the  presence  of  acid.  By  calculation, 
in  a  manner  similar  to  that  described  under  the  standard 
dichromate  solution,  it  is  found  necessary  to  have  one- 
tenth  of  the  2  gram-molecular  weights  of  potassium 
permanganate  (316.06  grams),  or  31.606  grams,  in  each 
liter  of  solution;  whereby  this  volume  of  solution  will  be 
capable  of  furnishing  just  sufficient  oxygen  to  oxidize 
1.01  grams  of  hydrogen.  The  change  from  the  deep 
purple  color  of  the  permanganate  to  the  almost  colorless 
manganous  salt  serves  admirably  to  mark  the  point  when 
the  reduction  is  complete  and  hence  to  fix  the  end-point 
in  titration  with  a  permanganate  solution. 

Example  69.  - —  100  c.c.  of  a  hydrogen  peroxide  solution  were 
required  to  decolorize  500  c.c.  of  N/10  potassium  permanganate 
solution  (acidulated).  What  was  the  percentage  concentration 
of  the  hydrogen  peroxide  solution? 

2  KMn04  +  3  H2SO4  +  5  H2O2 

=  K2S04  +  2  MnSO4  +  3  H2O  +  5  H20  +  5  O2. 

As  the  equation  indicates,  the  5  atoms  of  oxygen  liberated 
from  the  permanganate  withdraw  5  atoms  of  oxygen  from 
5  molecules  of  hydrogen  peroxide  and  are  together  removed 
from  the  reaction  in  the  molecular  form  (5  molecules). 
500  c.c.  of  N/10  potassium  permanganate  solution  are 
equivalent  to  50  c.c.  of  the  normal  solution.  One  liter  of 


COMPLEX   EQUATIONS  157 

a  normal  potassium  permanganate  solution,  containing 
31,606  grams  (1/10  of  316.06  grams  as  represented  by  the 
reaction-quantity),  can  only  react  with  1/10  of  the  quan- 
tity 5  H202  required  by  the  reaction,  or  with  1/2  of  one 
molecule  of  hydrogen  peroxide.  Here  50  c.c.  (1/20  liter) 
can  act  upon  just  1/20  of  the  amount  of  hydrogen  per- 
oxide that  a  liter  of  the  normal  solution  can  oxidize 
(namely,  1/2  of  a  gram-molecular  weight,  or  17  grams), 
i.e.,  1/20  of  17  grams  or  0.85  gram.  Consequently  in 
100  c.c.  of  the  hydrogen  peroxide  solution  there  is  only 
this  amount,  0.85  gram,  of  the  compound,  and  if  the 
weight  of  the  solution  is  assumed  to  be  100  grams,  then 
the  percentage  of  hydrogen  peroxide  present  is  85/100  of 

1  per  cent. 

From  the  conditions  of  this  reaction  the  amount  of  any 
one  substance  present  can  be  calculated  also  from  the 
volume  of  oxygen  evolved. 

Just  as  potassium  dichromate  so  also  an  acidulated 
solution  of  potassium  permanganate  readily  oxidizes  solu- 
tions of  ferrous  salts  to  the  ferric  state.  The  equation 
indicates,  of  course,  this  liberation  of  5  oxygen  atoms  per 

2  molecules  of  permanganate: 

2  KMnO4  +  10  FeS04  +  8  H2S04 

=  K2SO4  +  2  MnS04  +  5  Fe2(SO4)8  +  8  H20. 

Again  the  action  of  an  acidulated  potassium  perman- 
ganate solution  upon  a  hot  solution  of  oxalic  acid  (weighed 
out  as  C2H2O4 .  2  H2O),  with  the  complete  oxidation  of 
the  latter  into  carbon  dioxide  and  water,  affords  an 
admirable  means  for  the  titration  and  standardization  of 
permanganate  solutions : 

2  KMn04  +  5  C2H204 .  2  H20  +  3  H2S04 

=  K2S04  +  2  MnS04+  10  CO2  +  8  H2O  +  (10  H20). 


158  CHEMICAL   CALCULATIONS 

Hydrochloric  acid,  except  in  the  cold  and  under  proper 
precautions,  is  rarely  ever  used  for  the  acidulation  of 
permanganate  solutions  owing  to  its  ready  oxidation, 
similar  to  that  by  potassium  dichromate,  into  chlorine 
and  water: 

2  KMn04  +  16  HC1  =  2  KC1  +  2  MnCl2  +  5  C12  +  8  H20. 

In  neutral  or  alkaline  solutions  potassium  permanganate 
is  reduced  by  the  presence  of  oxidizable  substances  to 
manganese  dioxide  (precipitated),  and  somewhat  less 
oxygen  per  molecule  of  permanganate  is  thereby  liberated. 
The  action  may  be  represented  in  part  as  follows: 

2  KMnO4  +  H2O  =  2  KOH  +  2  Mn02  +  3  O. 

The  3  oxygen  atoms  liberated  from  2  molecules  of  potas- 
sium permanganate  determine  the  amount  of  this  salt 
in  its  normal  solution  to  be  1/6  of  316.06  grams  (repre- 
sented by  the  reaction-quantity  2  KMn04),  or  52.67 
grams  per  liter. 

lodimetry.  —  The  action  of  iodine  in  the  presence  of 
water  upon  a  number  of  compounds  is  interesting  to  show 
one  of  the  indirect  methods  of  estimating  substances  in 
solution. 

Iodine  dissolves  readily  in  water  containing  potassium 
iodide.  This  solution  acts  as  a  mild  oxidizing  agent  as 
may  be  seen  from  the  following  equation: 

S02  +  2  H20  +  I2  =  H2SO4  +  2  HI. 

In  the  presence  of  an  oxidizable  substance  the  one  mole- 
cule of  iodine,  through  the  decomposition  of  water  present, 
furnishes  sufficient  oxygen  to  oxidize  one  molecule  of 
sulphur  dioxide  to  the  trioxide,  or  an  amount  capable 
also  of  oxidizing  two  atoms  of  hydrogen.  The  normal 
value,  therefore,  of  such  an  iodine  solution  must  be  fixed 


COMPLEX   EQUATIONS  159 

at  one-half  the  gram-molecular  weight  (253.84  grams),  or 
126.92  grams  per  liter.  The  employment,  however,  of  a 
more  dilute  solution  as  N/10,  etc.,  is  more  common  in 
practice.  -The  end-point  in  titrations  with  iodine  solu- 
tions is  detected  by  the  addition  of  a  slight  amount  of  a 
starch  solution,  with  which  substance  the  iodine,  when 
occurring  in  slight  excess,  forms  a  deep  blue  mixture. 

An  iodine  solution  may  also  exert  an  oxidizing  action 
by  withdrawing  a  number  of  positively  charged  ions 
from  a  salt  in  solution. 

2  Na2S2O3  +  I2  =  Na2S4O6  +  2  Nal,   or 
4  Na  +  2  S2O3"  +  I2  =  2  Na*  +  S4Oe"  +  2  Na*  +  2  P. 

A  solution  of  sodium  thiosulphate  can  be  easily  made  up 
from  the  well  crystallized  salt,  Na2S2O3  .  5  H2O,  and  this 
solution  titrated  against  an  iodine  solution,  using  starch 
as  an  indicator.  One  molecule  of  iodine  reacts  with  two 
molecules  of  the  thiosulphate;  consequently  a  normal 
solution  of  the  latter  will  contain  (with  reference  to  a 
normal  solution  of  iodine)  just  one-half  of  this  reaction- 
quantity  or  one  gram-molecular  weight  per  liter. 

Substances  that  are  capable  of  liberating  iodine  from 
potassium  iodide,  e.g.  chlorine,  bromine,  etc.,  may  be 
brought  in  this  way  into  estimations  by  volumetric 
analysis : 

2  KI  +  C12  =  2  KC1  4-  I2. 

The  amount  of  iodine  liberated  is  in  exact  accord  with 
the  proportional  amount  of  chlorine  under  investigation, 
and  the  iodine  in  turn  is  open  to  determination  from  the 
amount  of  thiosulphate  required  for  titration.  This  in- 
teresting application  of  volumetric  analysis  is  known  as 
lodimetry. 

Example  70.  —  An  unknown  weight  of  sodium  dichromate 
was  warmed  with  an  excess  of  hydrochloric  acid,  and  the  chlorine, 


160  CHEMICAL   CALCULATIONS 

set  free,  passed  into  a  solution  of  potassium  iodide.  400  c.c.  of 
N/10  sodium  thiosulphate  solution  were  required  for  the  titra- 
tion  of  the  iodine  thereby  liberated.  Determine  the  amount  of 
chlorine  evolved,  and  also  the  weight  of  dichromate  reduced. 

400  c.c.  of  N/10  sodium  thiosulphate  solution  are 
equivalent  to  40  c.c.  of  a  normal  solution.  1  liter  of  a 
normal  solution  of  this  salt  exactly  decolorizes  1  liter  of 
a  normal  solution  of  iodine,  or  this  amount  of  iodine 
(126.92  grams)  in  whatever  volume  of  solution  it  is  con- 
tained. Consequently  40  c.c.  of  the  normal  thiosulphate 
solution  will  act  upon  that  amount  of  iodine  as  indi- 
cated in  the  proportion: 

1000  :  40  =  126.92  :  x,  or  x  =  5.177  grams. 

The  amount  of  chlorine  proportional  to  this  weight  of 
iodine  is  in  accord  with  the  ratio  Cl/I,  or  35.46/126.92, 
as  indicated  by  the  reaction  above.  By  proportion  we 
have:  126.92  :  35.46  =  5.177  :  x,  where  x,  or  1.45 
grams,  is  the  weight  of  chlorine  set  free  and  thus 
made  capable  of  liberating  the  amount  of  iodine  above 
calculated.  The  determination  of  the  amount  of  sodium 
dichromate  necessary  to  liberate  this  weight  of  chlorine 
rests  upon  an  equation  exactly  analogous  to  the  action 
of  hydrochloric  acid  upon  potassium  dichromate  (see 
p.  153),  and  is  indicated  by  the  ratio: 

3  a,        or  212.76 


Na2Cr2O7'          262 

Accordingly,  212.76  :  262  =  1.45  :  x.     From  this  we  cal- 
culate the  weight  of  dichromate  as  1.786  grams. 

No  matter  how  complex  the  equations  may  become,  the 
true  relationship  between  all  of  the  substances  concerned 
is  found  to  be  in  direct  proportionality  to  their  respec- 
tive reaction-quantities. 


COMPLEX   EQUATIONS  161 

PROBLEMS. 

357.  What  weight  of  ferric  chloride,  FeCl3,  can  be  reduced 
to  ferrous  chloride,  FeCl2,  by  8.52  grams  of  hydrogen  sulphide? 

Ans.  81.10  grams. 

358.  What  volume  of  hydrogen  sulphide  (at  standard  con- 
ditions) will  be  required  for  the  reduction  of  100  grams  of  ferric 
chloride,  FeCl3,  to  the  ferrous  salt?  Ans.  6903.8  c.c. 

259.  Calculate  the  weight  of  sulphur  precipitated  in  the 
reduction  of  100  grams  of  ferric  chloride,  FeCl3,  to  the  ferrous 
salt  by  the  action  of  hydrogen  sulphide.  Ans.  9.884  grams. 

360.  1000  c.c.  of  hydrogen  sulphide  (at  standard  conditions) 
will  reduce  what  weight  of  potassium  dichromate  in  acid  solu- 
tion? Ans.  4.378  grams. 

361.  What  volume  of  hydrogen  sulphide  (at  standard  con- 
ditions) will  be  required  for  the  reduction  of  100  grams  of  potas- 
sium dichromate  in  acid  solution?  Ans.  22,842  c.c. 

363.  What  weight  of  hydrogen  sulphide  can  be  oxidized  by 
400  grams  of  potassium  dichromate,  in  acid  solution?  What 
weight  of  sulphur  will  be  precipitated? 

Ans.  139.05  grams  HjS. 
130.81  grams  S. 

363.  What  volume  of  sulphur  dioxide  (at  standard  condi- 
tions) can  be  oxidized  by  30  grams  of  potassium  dichromate  in 
acid  solution?  Ans.  6852  c.c. 

364.  What  volume  of  hydrogen  sulphide,  at  24°  and  750  mm. 
pressure,  can  be  oxidized  by  10  grams  of  potassium  dichromate 
in  acid  solution?  Ans.  2518  c.c. 

365.  What  weight  of  potassium  dichromate,  in  acid  solution, 
will  be  reduced  by  1653.5  c.c.  of  sulphur  dioxide  at  22°  and 
745  mm.  pressure?  Ans.  6.567  grams. 

366.  Cakulate  the  volume  of  chlorine  liberated  in  the  action 
of  40  grams  of  potassium  dichromate  upon  a  hydrochloric  acid 
solution.  Ans.  9137  c.c. 

367.  What  weight  of  sodium  dichromate  must  enter  into 
reaction  with  a  hydrochloric  acid  solution  in  order  to  liberate 
100  grams  of  chlorine?  Ans.  123.14  grams. 


162  CHEMICAL  CALCULATIONS 

268.  Determine  the  normality  of  a  potassium  dichromate 
solution,  25  c.c.  of  which  oxidized  1.24  grams  of  ferrous  sulphate 
to  the  ferric  salt.  Ans.  0.3265  N. 

269.  Determine  the  normality  of  a  sodium  dichromate  solu- 
tion, 50  c.c.  of  which  oxidized  3.85  grams  of  ferrous  ammonium 
sulphate,  FeS04 .  (NH4)2S04 .  6  H2O  to  the  ferric  salt. 

Ans.  0.1963  N. 

270.  What  volume  of  hydrogen  sulphide,  at  standard  con- 
ditions, will  be  required  to  reduce  200  c.c.  of  N/10  potassium 
dichromate  solution  (acidulated)?  Ans.  224  c.c. 

271.  What  weight  of  potassium  permanganate,  in  acid  solu- 
tion, will  be  reduced  by  5000  c.c.  of  hydrogen  sulphide  at 
standard  conditions?  Ans.  14.11  grams. 

272.  Calculate  the  volume  of  hydrogen  sulphide,  at  stand- 
ard conditions,  that  can  be  oxidized  by  4  grams  of  potassium 
permanganate  in  acid  solution.  Ans.  1417.4  c.c. 

273.  What  weight  of  sulphur  dioxide  can  be  oxidized  by 
200  grams  of  potassium  permanganate  in  acid  solution? 

Ans.  202.7  grams. 

274.  50  c.c.  of  a  solution  of  hydrogen  peroxide  were  required 
to  decolorize  400  c.c.  of  N/5  potassium  permanganate  solution 
(acidulated).     Calculate   the  percentage  concentration  of  the 
hydrogen  peroxide  solution.  Ans.  2.72  per  cent. 

275.  Calculate  the  weight  of  crystallized  oxalic  acid,  C2H204 . 
2  H20,  required  for  the  reduction  of  100  grams  of  potassium 
permanganate  in  acid  solution.    What  volume  of  carbon  diox- 
ide, at  standard  conditions,  would  be  liberated? 

Ans.  199.4  grams. 
70,873  c.c.  CO,. 

276.  50  c.c.  of  an  acidulated  potassium  permanganate  solu- 
tion were  reduced  by  2.4  grams  of  anhydrous  oxalic  acid, 
C2H2O4.     Calculate  the  normality  of  the  permanganate  solution. 

Ans.  1.066  N. 

277.  What  weight  of  sulphur  dioxide  will  be  oxidized  by  50 
grams  of  potassium  permanganate  in  alkaline  solution? 

Ans.  3.04  grams. 


COMPLEX   EQUATIONS  163 

278.  The   bromine   set   free   by  the   action  of  manganese 
dioxide  upon  a  hydrobromic  acid  solution  was  passed  into  a 
solution  of  potassium  iodide.     200  c.c.  of  N/10  sodium  thio- 
sulphate  solution  were  required  for  the  titration  of  the  free 
iodine.     Calculate  the  weight  of  bromine  evolved. 

Ans.  1.6  grams. 

279.  Determine  the  purity  of  a  sample  of  manganese  dioxide, 
2.2  grams  of  which,  with  excess  of  hydrochloric  acid,  set  free 
sufficient  chlorine  to  liberate  a  quantity  of  iodine  that  required 
250  c.c.  of  N/5  sodium  thiosulphate  solution  for  titration. 

Ans.  98.79  per  cent  pure. 

280.  0.2452  gram  of    potassium  dichromate,   acting   upon 
an  excess  of  hydrochloric  acid,  set  free  an  amount  of  chlorine 
which,  when  passed  into  a  solution  of  potassium  iodide,  liber- 
ated iodine  sufficient  for  the  titration  of  50  c.c.  of  an  unknown 
sodium  thiosulphate  solution.     Calculate  the  normality  of  this 
latter  solution.  .Ans.  N/10. 


APPENDIX. 


APPENDIX  I. 
Correction  of  Barometic  Readings. 

A  barometric  reading  at  room  temperature  is  reduced 
to  the  corresponding  height  of  a  column  of  mercury  at  0° 
by  subtracting  from  the  actual  reading  in  millimeters 
that  number  in  the  second  column  below  (Correction)  set 
opposite  the  observed  temperature.  The  barometric  read- 
ings throughout  these  problems  are  given  in  corrected 
form. 


Tempera- 

Correc- 

Tempera- 

Correc- 

Tempera- 

Correc- 

ture. 

tion. 

ture. 

tion. 

ture. 

tion. 

12 

1.6 

17 

2.2 

23 

3.0 

13 

1.7 

18.5 

2.4 

24.5 

3.2 

14 

1.8 

20 

2.6 

25 

3.3 

15 

2.0 

21.5 

2.8 

26 

3.4 

APPENDIX   II. 
Tension  of  Aqueous  Vapor  in  Millimeters. 


Tempera- 

Pressure. 

Tempera- 

Pressure. 

Tempera- 

Pressure. 

ture. 

ture. 

ture. 

0° 

4.6 

16° 

13.5 

26° 

25.1 

5 

6.5 

17 

14.4 

27 

26.5 

8 

8.0 

ia 

15.  .4 

2S 

28.1 

9 

8.6 

19 

16.3 

29 

29.8 

10 

9.2- 

20 

17.4 

30 

31.5 

11 

9.8 

21 

18.5 

31 

33.4 

12 

10.5 

22 

19.7 

32 

35.4 

13 

11.2 

23 

20.9 

33 

37.4 

14 

11.9 

24 

22.2 

34 

39.6 

15 

12.7 

25 

23.6 

100 

760.0 

167 


168 


CHEMICAL   CALCULATIONS 


APPENDIX  III.  — Four-Figure  Logarithms. 


irst  two 
igures. 

Third  figure. 

Fourth-figure 
differences. 

70         f\ 

tn~ 

0334 

o    9 

10 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0374 

4    8  12 

17  21  25 

29  33  37 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

4    8  11 

15  19  23 

26  30  34 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

3    7  10 

14  17  21 

24  28  31 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

3    6  10 

13  16  19 

23  26  29 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

,369 

12  15  18 

21  24  27 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

368 

11  14  17 

20  22  25 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

358 

11  13  16 

18  21  24 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

2    5    7 

10  12  15 

17  20  22 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

257 

9  12  14 

16  19  21 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

2    4    7 

9  11  13 

16  18  20 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

2         6 

a  11  13 

15  17  19 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

2         0 

8  10  12 

14  16  18 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

2         6 

8  10  12 

14  15  17 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

2         6 

7    9  11 

13  15  17 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

2         5 

7    911 

12  14  16 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

2    3 

7    9  10 

12  14  15 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

2    3 

7    8  10 

11  13  15 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

2    3 

689 

11  13  14 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

2    3 

689 

11  12  14 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

1    3 

679 

10  12  13 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

3 

679 

10  11  13 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

3 

678 

10  11  12 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

3 

578 

9  11  12 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

3 

568 

9  10  12 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

3 

568 

9  10  11 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

2 

567 

9  10  11 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

2 

567 

8  10  11 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

2 

567 

8    9  10 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

2 

567 

8    9  10 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

2 

457 

8    9  10 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

2 

456 

8    9  10 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

2 

456 

789 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

2 

5    6 

789 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

2 

5    6 

789 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

2 

5    6 

789 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

2 

5    6 

789 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

2 

5    6 

778 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

2 

5    5 

678 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

2 

445 

678 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

2 

445 

678 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

2 

345 

678 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

2 

345 

678 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

2 

345 

677 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

2 

345 

667 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

2 

345 

667 

APPENDIX 


169 


Four-Figure  Logarithms. 


rst  two 
gures. 

Third  figure. 

Fourth-figure 
differences. 

E« 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

123 

456 

789 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

122 

345 

567 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

1    2    2 

345 

567 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

122 

345 

567 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

1    1    2 

344 

567 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

1    1    2 

344 

567 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

1         2 

344 

566 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

1         2 

344 

566 

62 
63 

7924 
7993 

7931 
8000 

7938 

son? 

7945 
8014 

7952  7959 
8021  8028 

7966 
8035 

7973 
8041 

7980 
8048 

7987 
8055 

1         2 
1         2 

334 
334 

566 
556 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

1         2 

334 

556 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

^112 

334 

556 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

1    1    2 

334 

556 

67 

S:M1 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

112 

334 

556 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

1    1    2 

334 

456 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

1    1    2 

234 

456 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

1         2 

234 

456 

71 

8513 

8519 

S525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

1         2 

234 

455 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609- 

8615 

8621 

8627 

1         2 

234 

455 

73 

8633 

SB39 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

1         2 

234 

455 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

1         2 

234 

455 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

1         2 

233 

455 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

1         2 

233 

455 

77 

8865 

ss7! 

S87t> 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

1         2 

233 

4         5 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

1         2 

233 

4         5 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

1         2 

233 

4         5 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

1         2 

233 

4         5 

81 

UOS5 

1090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

1         2 

233 

4         5 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

1         2 

233 

4         5 

83 

9191 

n% 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

1         2 

233 

4         5 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

1         2 

20        Q 
O      O 

4         5 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

1         2 

233 

4         5 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

1         2 

233 

4         5 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

0 

223 

3         4 

88 

9445 

.)}.-,() 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

0 

223 

3         4 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

0 

223 

3         4 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

0 

223 

3         4 

91 

9590 

9595 

9600 

9605 

9609 

%14 

9619 

9624 

9628 

9633 

0 

223 

3         4 

92 

9MX 

9643 

9647 

<)C,.-,L> 

9657 

i«i(n 

9666 

9671 

9675 

imxo 

0 

223 

344 

93 

9685 

9689 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

0 

223 

3    4 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

0 

223 

3    4 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

0 

223 

3    4 

96 

9823 

9827 

»s:w 

i)s:-;i; 

9841 

9845 

<)x;,o 

9854 

JSf,H 

9863 

0 

223 

3    4 

97 

9868 

9872 

9S77 

068] 

'issii 

iisuo 

9894 

9899 

.)<)(>:; 

9908 

0 

223 

3    4 

98 

9912 

9917 

9921 

•):)2t; 

9930 

9934 

9939 

9943 

9948 

9952 

0 

223 

3    4 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

0 

223 

3    3 

170  CHEMICAL   CALCULATIONS 

APPENDIX  IV.  —  Antilogarithms. 


s^i 

*gf 

Third  figure. 

Fourth-figure 
differences. 

S3| 

E«B 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

123 

456 

789 

.00 

1000 

1002 

1005 

1007 

1009 

1012 

1014 

1016 

1019 

1021 

0    0    1 

1    1    1 

2    2 

.01 

1023 

1026 

1028 

1030 

1033 

1035 

1038 

1040 

1042 

1045 

0    0    1 

1    1    1 

2    2 

.02 

1047 

1050 

1052 

1054 

1057 

1059 

1062 

1064 

1067 

1069 

0    0    1 

1    1    1 

2    2 

.03 

1072 

1074 

1076 

1079 

1081 

1084 

1086 

1089 

1091 

1094 

0    0    1 

1    1    1 

2    2 

.04 

1096 

1099 

1102 

1104 

1107 

1109 

1112 

1114 

1117 

1119 

0    1    1 

1    1    2 

2    2 

.05 

1122 

1125 

1127 

1130 

1132 

1135 

1138 

1140 

1143 

1146 

0         1 

2 

2    2 

.06 

1148 

1151 

1153 

1156 

1159 

1161 

1164 

1167 

1169 

1172 

0         1 

5 

2    2 

.07 

1175 

1178 

1180 

1183 

1186 

1189 

1191 

1194 

1197 

1199 

0         1 

2    2 

.08 

1202 

1205 

1208 

1211 

1213 

1216 

1219 

1222 

1225 

1227 

0         1 

2    2 

.09 

1230 

1233 

1236 

1239 

1242 

1245 

1247 

1250 

1253 

1256 

0         1 

2    2 

.10 

1259 

1262 

1265 

1268 

1271 

1274 

1276 

1279 

1282 

1285 

0    1    1 

1 

2    2 

.11 

1288 

1291 

1294 

1297 

1300 

1303 

1306 

1309 

1312 

mr 

0    1    1 

2 

2    2 

.12 

1318 

1321 

1324 

1327 

1330 

1334 

1337 

1340 

1343 

1346 

Oil 

2 

2    2 

.13 

1349 

1352 

1355 

1358 

1361 

1365 

1368 

1371 

1374 

1377 

Oil 

2 

2    3 

.14 

1380 

1384 

1387 

1390 

1393 

1396 

1400 

1403 

1406 

1409 

Oil 

2 

2    3 

.15 

1413 

1416 

1419 

1422 

1426 

1429 

1432 

1435 

1439 

1442 

Oil 

2 

233 

.16 

1445 

1449 

1452 

1455 

1459 

1462 

1466 

1469 

1472 

1476 

0    1     1 

2 

233 

.17 

1479 

1483 

1486 

1489 

1493 

1496 

1500 

1503 

1507 

1510 

0    1 

2 

233 

.18 

1514 

1517 

1521 

1524 

1528 

1531 

1535 

1538 

1542 

1545 

0    1 

2    2 

233 

.19 

1549 

1552 

1556 

1560 

1563 

1567 

1570 

1574 

1578 

1581 

0    1 

2    2 

333 

.20 

1585 

1589 

1592 

1596 

1600 

1603 

1607 

1611 

1614 

1618 

0    1 

1    2    2 

333 

.21 

1622 

1626 

1629 

1633 

1637 

1641 

1644 

1648 

1652 

1656 

0    I 

222 

333 

.22 

1660 

1663 

1667 

1671 

1675 

1679 

1683 

1687 

1690 

1694 

0    1    1 

222 

333 

.23 

1698 

1702 

1706 

1710 

1714 

1718 

1722 

1726 

1730 

1734 

0    1    1 

222 

334 

.24 

1738 

1742 

1746 

1750 

1754 

1758 

1762 

1766 

1770 

1774 

0    1    1 

222 

334 

.25 

1778 

1782 

1786 

1791 

1795 

1799 

1803 

1807 

1811 

1816 

0    1    1 

222 

334 

.26 

1820 

1824 

1828 

1832 

1837 

1841 

1845 

1849 

1854 

1858 

0    1    1 

223 

334 

.27 

1862 

1866 

1871 

1875 

1879 

1884 

1888 

1892 

1897 

1901 

0    1    1 

223 

334 

.28 

1905 

1910 

1914 

1919 

1923 

1928 

1932 

1936 

1941 

1945 

0    1    1 

223 

344 

.29 

1950 

1954 

1959 

1963 

1968 

1972 

1977 

1982 

1986 

1991 

0    1    1 

223 

344 

.30 

1995 

2000 

2004 

2009 

2014 

2018 

2023 

2028 

2032 

2037 

0    1    1 

223 

344 

.31 

2042 

2046 

2051 

2056 

2061 

2065 

2070 

2075 

2080 

2084 

0    1    1 

223 

344 

.32 

2089 

2094 

2099 

2104 

2109 

2113 

2118 

2123 

2128 

2133 

0    1    1 

223 

344 

.33 

2138 

2143 

2148 

2153 

2158 

2163 

2168 

2173 

2178 

2183 

0    1    1 

223 

344 

.34 

2188 

2193 

2198 

2203 

2208 

2213 

2218 

2223 

2228 

2234 

1    1    2 

233 

445 

.35 

2239 

2244 

2249 

2254 

2259 

2265 

2270 

2275 

2280 

2286 

1    1    2 

233 

445 

.36 

2291 

2296 

2301 

2307 

2312 

2317 

2323 

2328 

2333 

2339 

1    1    2 

233 

445 

.37 

2344 

2350 

2355 

2360 

2366 

2371 

2377 

2382 

2388 

2393 

1    1    2 

233 

445 

.38 

2399 

2404 

2410 

2415 

2421 

2427 

2432 

2438 

2443 

2449 

1    2 

233 

445 

.39 

2455 

2460 

2466 

2472 

2477 

2483 

2489 

2495 

2500 

2506 

1    2 

233 

455 

.40 

2512 

2518 

2523 

2529 

2535 

2541 

2547 

2553 

2559 

2564 

1    2 

234 

455 

.41 

2570 

2576 

2582 

2588 

2594 

2600 

2606 

2612 

2618 

2624 

1    2 

234 

455 

.42 

2630 

2636 

2642 

2649 

2655 

2661 

2667 

2673 

2679 

2685 

1    2 

234 

456 

.43 

2692 

2698 

2704 

2710 

2716 

2723 

2729 

2735 

2742 

2748 

1    2 

334 

456 

.44 

2754 

2761 

2767 

2773 

2780 

2786 

2793 

2799 

2805 

2812 

1    2 

334 

456 

.45 

2818 

2825 

2831 

2838 

2844 

2851 

2858 

2864 

2871 

2877 

1    2 

334 

556 

.46 

2884 

2891 

2897 

2904 

2911 

2917 

2924 

2931 

2938 

2944 

1    2 

334 

556 

.47 

2951 

2958 

2965 

2972 

2979 

2985 

2992 

2999 

3006 

3013 

1    2 

334 

556 

48 

3020 

3027 

3034 

3041 

3048 

3055 

3062 

3069 

3076 

3083 

1    2 

344 

566 

.49 

3090 

3097 

3105 

3112 

3119 

3126 

3133 

3141 

3148 

3155 

1    2 

344 

566 

APPENDIX 


171 


Antilogarithms. 


First  two 
figures  of 
mantissa. 

Third  figure. 

Fourth-figure 
differences. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

123 

456 

789 

.50 

3162 

3170 

3177 

3184 

3192 

3199 

3206 

3214 

3221 

3228 

1  2 

344 

567 

.51 

3236 

3243 

3251 

3258 

3266 

3273 

3281 

3289 

3296 

3304 

2  2 

345 

567 

.52 

3311 

3319 

3327 

3334 

3342 

3350 

3357 

3365 

3373 

3381 

2  2 

345 

567 

.53 

3388 

3396 

3404 

3412 

3420 

3428 

3436 

3443 

3451 

3459 

2  2 

345 

667 

.54 

3467 

3475 

3483 

3491 

3499 

3508 

3516 

3524 

3532 

3540 

2  2 

345 

667 

.55 

3548 

3556 

3565 

3573 

3581 

3589 

3597 

3606 

3614 

3622 

2  2 

345 

677 

.56 

3631 

3639 

3648 

3656 

3664 

3673 

3681 

3690 

3698 

3707 

2  3 

345 

678 

.57 

3715 

3724 

3733 

3741 

3750 

3758 

3767 

3776 

3784 

3793 

2  3 

345 

678 

.58 

3802 

3811 

3819 

3828 

3837 

3846 

3855 

3864 

3873 

3882 

2  3 

445 

678 

.59 

3890 

3899 

3908 

3917 

3926 

3936 

3945 

3954 

3963 

3972 

2  3 

455 

678 

.60 

3981 

3990 

3999 

4009 

4018 

4027 

4036 

4046 

4055 

4064 

1  2  3 

456 

678 

.61 

4074 

4083 

4093 

4102 

4111 

4121 

4130 

4140 

4150 

4159 

1  2  3 

456 

789 

.62 

4169 

4178 

4188 

4198 

4207 

4217 

4227 

4236 

4246 

4256 

123 

456 

789 

.63 

4266 

4276 

4285 

4295 

4305 

4315 

4325 

4335 

4345 

4355 

1  2  3 

456 

789 

.64 

4365 

4375 

4385 

4395 

4406 

4416 

4426 

4436 

4446 

4457 

1  2  3 

456 

789 

.65 

4467 

4477 

4487 

4498 

4508 

4519 

4529 

4539 

4550 

4560 

1  2  3 

456 

789 

.66 

4571 

4581 

4592 

4603 

4613 

4624 

4634 

4645 

4656 

4667 

1  2  3 

456 

7  9  10 

.67 

4677 

4688 

4699 

4710 

4721 

4732 

4742 

4753 

4764 

4775 

2  3 

457 

8  9  10 

.68 

4786 

4797 

1808 

4819 

4831 

4842 

4853 

4864 

4875 

4887 

2  3 

4.67 

8  9  10 

.69 

4898 

4909 

4920 

4932 

4943 

4955 

4966 

4977 

4989 

5000 

2  3 

567 

8  9  10 

.70 

5012 

5023 

5035 

5047 

5058 

5070 

5082 

5093 

5105 

5117 

2  4 

567 

8  9  11 

.71 

5129 

5140 

5152 

5164 

5176 

5188 

5200 

5212 

5224 

5236 

2  4 

567 

8  10  11 

.72 

5248 

5260 

5272 

5284 

7297 

5309 

5321 

5333 

5346 

5358 

2  4 

567 

9  10  11 

.73 

5370 

5383 

5395 

5408 

5420 

5433 

5445 

5458 

5470 

5483 

3  4 

568 

9  10  11 

.74 

5495 

5508 

5521 

5534 

5546 

5559 

5572 

5585 

5598 

5610 

3  4 

568 

9  10  12 

.75 

5623 

5636 

5649 

5662 

5675 

5689 

5702 

5715 

5728 

5741 

1  3  4 

578 

9  10  12 

.76 

5754 

5768 

5781 

5794 

5808 

5821 

5834 

5848 

5861 

5875 

134 

578 

9  11  12 

.77 

5888 

6902 

5916 

5929 

5943 

5957 

5970 

5984 

5998 

6012 

1  3  4 

578 

10  11  12 

.78 

6026 

6039 

6053 

6067 

6081 

6095 

6109 

6124 

6138 

6152 

1  3  4 

678 

10  11  13 

.79 

6166 

6180 

6194 

6209 

6223 

6237 

6252 

6266 

6281 

6295 

1  3  4 

679 

10  11  13 

.80 

6310 

6324 

6339 

6353 

6368 

6383 

6397 

6412 

6427 

6442 

1  3  4 

679 

10  12  13 

.81 

6457 

6471 

6486 

6501 

6516 

6531 

6546 

6561 

6577 

6592 

235 

689 

11  12  14 

.82 

6607 

6622 

6637 

6653 

6668 

6683 

6699 

6714 

6730 

6745 

235 

689 

11  12  14 

.83 

6761 

6776 

6792 

6808 

6823 

0839 

6855 

:cs7i 

6887 

6902 

235 

689 

11  13  14 

.84 

6918 

6934 

6950 

6966 

6982 

6998 

7015 

7031 

7047 

7063 

235 

6  8  10 

11  13  15 

.85 

7079 

7096 

7112 

7129 

7145 

7161 

7178 

7194 

7211 

7228 

235 

7  8  10 

12  13  15 

.86 

7244 

7261 

7278 

7295 

7311 

7328 

7345 

7362 

7379 

7396 

235 

7  8  10 

12  13  15 

.87 

741:5 

7430 

7447 

7464 

7482 

7499 

7516 

7534 

7551 

7568 

235 

7  9  10 

12  14  16 

.88 

7586 

7603 

7621 

7638 

705« 

7674 

7691 

7709 

7727 

7745 

245 

7  9  11 

12  14  16 

.89 

7762 

7780 

7798 

7816 

7834 

7852 

7870 

7889 

7907 

7925 

245 

7  9  11 

13  14  16 

.90 

7943 

7962 

7980 

7998 

8017 

8035 

8054 

8072 

8091 

8110 

246 

7  9  11 

13  15  17 

.91 

8128 

8147 

8166 

8185 

S204 

8222 

8241 

8260 

8279 

8299 

246 

8  9  11 

13  15  17 

.92 

8318 

8337 

8356 

8375 

8395 

8414 

8433 

8453 

8472 

5492 

2    6 

8  10  12 

14  15  17 

.93 

8511 

8531 

8551 

8570 

8590 

8610 

8630 

8650 

8670 

8690 

2    6 

8  10  12 

14  16  18 

.94 

8710 

8730 

8750 

8770 

8790 

8810 

8831 

8851 

8872 

8892 

2    6 

8  10  12 

14  16  18 

.95 

8913 

8933 

8954 

8974 

8995 

9016 

9036 

9057 

9078 

9099 

2    6 

8  10  12 

15  17  19 

.96 

9120 

9141 

9162 

91  W 

9204 

l.'L't', 

9247 

<)2f,S 

9290 

9311 

2    6 

8  11  13 

15  17  19 

.97 

9333 

9354 

9376 

9397 

9419  i  9441 

9462 

9484 

9506 

9528 

247 

9  11  13 

15  17  20 

.98 

9550 

9572 

9594 

9616 

9638  j  9661 

9683 

9705 

9727 

9750 

247 

9  11  13 

16  18  20 

.99 

9772 

9795  0317 

9840 

9863  9886 

9908 

9931 

9954 

9977 

2  5  7 

9  11  14 

16  18  20 

INDEX 


Analysis,     volumetric     and 

gravimetric 118 

Aqueous  vapor,  tension  of . .  .  28 

Atom,  definition  of 35 

Atomic  weight,  definition  of,  36,  46 

Avogadro's  hypothesis 33 

Boiling-point 29 

Boyle,  law  of 9 


Charles,  law  of 

Coefficient   of   expansion   of 


Cohesion  in  gases 


15 

15 
34 


Dalton,  law  of 27 

Definite  proportions,   law  of  46 
Density,  absolute,  definition 

of 5 

relative,     calculation      of, 

from  molecular  weight  42 

definition  of  „ 5 

standard  of,  for  gases. .  6 

upon  different  standards  38 
relation  of,  to  molecular 

weight 34 

to  'pressure  of  gases. ...  12 
to  temperature  of  gases  18 
to  pressure  and  tempera- 
ture of  gases 26 

Dyne,  definition  of 2 

Empirical  formulae,  definition 

of 61 

derivation  of,  from  percent- 
age composition 60 

End-point  in  titrations 106 

Erg,  definition  of    3 

Equations,  dependent 92 

independent 94 

study  of 82 

volume  relations  in .  .  86 


PAGE 
Force,  unit  of 2 

Formulae,  derivation  of 59 

for  salts  containing  water 

of  crystallization.  ...        70 

from  percentage  composi- 
tion   60,65 

from  relation  of  formula- 
quantities  to  each 
other 67 

from  relation  between  for- 
mula-quantities and 
their  corresponding 
weights 71 

from  relation  between  for- 
mula-quantities and 
percentage  composi- 
tion   74 

from  single  analytical  val- 
ues    75 

Formula-quantity,  definition 

of 48 

relation   of,   to   molecular 

weight 49 

to  other  formula-quan- 
tities  51,53 

to  percentage  composi- 
tion         52 

Gay-Lussac,  law  of 33 

Gram,  definition  of 2 

Gram-molecular  volume,  defi- 
nition of 40 

in  chemical  reactions 122 

Gram-molecular  weight,  defi- 
nition of 37 

Gravimetric  analysis 118 

Indicators  in  titrations 106 

lodimetry 158 

Ions,  hydrogen  and  hydrox- 
ide   106 


173 


174 


INDEX 


Length,  unit  of. . .% . 
Liter,  definition  of. 


PAGE 
1 
2 


Mass,  definition  of 3 

unit  of 2 

Meter,  definition  of 1 

Molar  solution,  definition  of  107 

Molecule,  definition  of 33 

Molecular  formulae,  definition 

of 46,61 

derivation  of,  from  percent- 
age composition 59 

Molecular  volumes  of  gases, 

combinations  between  122 

relation  of,  to  volume-unit  124 
Molecular  weight,  definition 

of 34,46 

Neutralization 106 

Normality  factor 109 

calculation  of,    by  simple 

proportion Ill 

Normal  solution,  definition  of  108 

Oxidation,  definition  of.  ...      148 
Oxidizing     solutions,      with 

KgCrjjOy 150 

withKMnO4 154 

Oxygen,  as  standard  of  rela- 
tive densities  of  gases .          6 
as  standard  of  molecular 

weights 6 

Partial  pressures  of  gases  ...        27 
Percentage    composition    of 

compounds 46 

Percentage  composition,  re- 
lation of,  to  formula- 
quantities  52 

as     basis    for    percentage 

purity 53 

Percentage  purity 53 

calculation  of 88 

Pressure,  normal 3 

relation  of,  to  temperature 

of  gases 23 

to  volume  of  gases.  .  . 
to  volume  and  tempera- 
ture of  gases 26 

Pressure-fraction 11 

Products  resulting  from  mix- 
tures, calculation  of . .  89 


PAGE 

Reaction-quantities 82 

calculation    of    ratios  be- 
tween   92, 94 

calculation  of,  from  weights 

of  substances  involved       97 

complex 98 

Reduction,  definition  of  ....      148 

Second,  definition  of 1 

Solutions,  adjustment  of,  to 

desired  standard 117 

comparison  of,  with  stand- 
ard       116 

standard 108 

standard  oxidizing 153, 156 

Specific  gravity,  definition  of         5 
Standardization  of  solutions, 

112,113 

Temperature,  normal 3 

relation  of,  to  density  of 


to  pressure  of  gases .... 

to  volume  of  gases 

to  volume  and   pressure 


18 
23 
18 

25 

Temperature-fraction 18 

Time,  unit  of 1 

Titration  of  solutions 106 

Valence,  definition  of 148 

Vapor  density 6 

Volume  relations  in  chemical 

equations 86 

Volume,  relation  of,  to  pres- 
sure of  gases 9 

to  temperature  of  gases        17 
to  temperature  and  pres- 
sure of  gases 21,  24 

unit  of 2 

Volume-fraction 11 

Volume-unit,    for    reactions 

between  gases 124 

calculation  of,  from  known 
volumes      of      gases ; 

125,126,  128 

calculation    of,    from    ob- 
served volume-changes     129 
calculation  of,    from  sev- 
eral simultaneous  re- 
actions     133, 135 

Volumetric  analysis 118 

Work,  unit  of 3 


BERKELEY  LIBRARIES 


THI 


INTERNATIONAL  ATOMIC  WEIGHTS  FOR  1917* 
O  =  16.00 


Name. 

Symbol. 

Atomic 
Weight. 

Name. 

Symbol. 

Atomic 
Weight. 

Aluminium  
Antimony  ....... 

Al    - 
Sb  - 

27.1 
120.2 

Molybdenum  
Neodymium  

Mo 
Nd 

96.0 
144.3 

A 

39  88 

Neon  i  ....... 

Ne 

20  2 

As  . 

74  96 

Nickel 

Ni 

58  68 

Ba  - 

137.37 

Niton  

Nt 

222  4 

Bi 

208.0 

N  — 

14  01 

B 

11  0 

Osmium 

Os 

190  9 

Br 

79  92 

Oxygen 

O   - 

16  00 

Cadmium 

Cd 

112.40 

Palladium  

Pd 

106  7 

Caesium 

Cs 

132.81 

Phosphorus  

P 

31  04 

Ca  - 

40  07 

Platinum 

Pt 

195  2 

C    . 

12  005 

Potassium 

K  ^ 

39  10 

Cerium  
Chlorine 

Ce 
Cl 

140.25 
35.46 

Praseodymium... 
Radium  

Pr 
Ra 

140.9 
226  0 

Chromium 

Cr 

52  0 

Rhodium 

Rh 

102  9 

Cobalt  
/olumbium  
>pper  *-.».... 
)yspros5\im  
Vjium 

Co 
Cb 
Cu- 
Dy 
Er 

58.97 
93.1 
63.57 
162.5 
167.7 

Rubidium  
Ruthenium  
Samarium  
Scandium  
Selenium  ... 

Rb 
Ru 

Sa 
Sc 
Se  -. 

85.45 
101.7 
150.4 
44.1 
79  2 

ropium 

Eu 

152  0 

Silicon 

Si  — 

28  3 

uorine  

F  - 

19.0 

Silver  

Ag  . 

107  88 

iadolinium  
!ul  1  i  ui  n  

Gd 

Ga 

157.3 
69.9 

Sodium  
Strontium  

Na   <• 
Sr 

23.00 
87  63 

lermanium  .  .   . 

Ge 

72.5 

Sulphur 

S      ~ 

32  06 

Huciuuni 

Gl 

9  1 

Tantalum 

Ta 

181  5 

Jold  

Au 

197.2 

Tellurium  

Te 

127.5 

lelium  
lolmium  
ly  drogen  

He 
Ho 
H 

4.00 
163.5 
1  008 

Terbium  
Thallium........ 
Thorium 

Tb 

Tl    « 
Th 

159.2 
204.0 
232  4 

ndium 

In 

114  8 

Thulium 

Tm 

168  5 

odine  ... 

I 

126  92 

Tin 

Sn 

118  7 

ridium 

Ir 

193  1 

Ti 

48  1 

ron 

Fe  - 

55  84 

W 

184  0 

Cry  pton  

Kr 

82  92 

Uranium 

u 

238  2 

/anthanuni 

La 

139  0 

Vanadium 

v 

51  0 

,ead  .   .   . 

Pb  - 

207  20 

Xe 

130  2 

.lithium 

Li 

6  94 

Yb 

173  5 

..utecium  

Lu 

175  0 

(  Neoy  tter  bium  ) 

lagnesium  
fanganese  

Mg 
Mn 

24.32 
54  93 

Yttrium  
Zinc 

Yt 
Zn    -. 

88.7 
65  37 

lercury  

Hg 

200  6 

Zirconium 

Zr 

90  6 

*  Compiled  by  the  International  Committee  on  Atomic  Weights  consisting  of 
W.  Clarke,  T.  E.  Thorpe,  and  G.  Urbain.    J.  Am.  Chem.  Soc.,  38,  2220. 


